Inequality Proof Problems [01-50]
Recommended posting: 【Algebra】 Algebra Index
P1. (IneqMath) Let $a, b, c > 0$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a+b+c}{\sqrt[3]{a b c}}+\frac{8 a b c}{(a+b)(b+c)(c+a)} \geq C.\]S1. $C = 4$
Let $t=\sqrt[3]{abc}$ and define $x=\frac{a}{t}, \quad y=\frac{b}{t}, \quad z=\frac{c}{t}.$ Then the left-hand side becomes
\[(x+y+z)+\frac{8}{(x+y)(y+z)(z+x)}\]where $xyz=1.$ Let $p=x+y+z, \quad q=xy+yz+zx.$ Using Schur’s inequality,
\[p^3+9xyz=p^3+9\ge 4pq,\]we can prove the desired statement.
P2. (IneqMath) For $a, b, c > 0$, find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a^3}{a^3 + abc + b^3} + \frac{b^3}{b^3 + abc + c^3} + \frac{c^3}{c^3 + abc + a^3} \geq C.\]S2. $C = 1$
The final inequality is established by the rearrangement inequality.
P3. (IneqMath) Find the smallest constant $C$ such that for all real numbers $x$ and $y$, the following inequality holds:
\[x^2 + x + y^2 + y + C \geq x y\]S3. $C = 1$
$f(x, y) = x^2 + x + y^2 + 1 - xy$ is differentiable at every point and does not diverge to negative infinity. Therefore, the minimum value of $f(x, y)$ is attained at a critical point. Setting
\[\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0\]gives the critical point $(-1, -1)$. At this point, the function value is $0$. Hence, the given inequality is proved.
P4. (IneqMath) Let $a, b, c \neq 0$ such that $a^2+b^2+c^2=2(ab+bc+ca)$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given condition:
\[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq C\]S4. $C = \frac{27}{2}$
Assume one of $a,b,c$ is negative. WLOG, let $c<0$. Since
\[a^2+b^2\ge 2ab,\qquad c^2>0>2(bc+ca),\]equality cannot hold. Similarly, if two of them are negative, equality also cannot hold. Thus $(a,b,c)$ have the same sign, so WLOG assume they are all positive. From the condition, $c=(\sqrt a+\sqrt b)^2.$ Let $a=\alpha^2,\qquad b=\beta^2=k\alpha^2,\qquad c=(\alpha+\beta)^2.$ Then the expression becomes
\[\left(1+k^2+(k+1)^2\right) \left(1+\frac1{k^2}+\frac1{(k+1)^2}\right).\]Its minimum occurs at $k=-2,\ -\frac12,\ 1,$ and the minimum value is $\frac{27}{2}.$ Therefore, $\text{the expression}\ge \frac{27}{2}.$
P5. (IneqMath) Let $a, b, c > 0$ such that $a + b + c = abc$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a}{b^3} + \frac{b}{c^3} + \frac{c}{a^3} \geq C.\]S5. $C = 1$
Let $p = a+b+c = abc,\qquad q = ab+bc+ca.$ Then we have
\[\left(\frac{q}{p}\right)^2=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\geq 3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=3.\]Therefore,
\[\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geq\frac{a}{a^3}+\frac{b}{b^3}+\frac{c}{c^3}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{q^2-2p^2}{p^2}= \left(\frac{q}{p}\right)^2-2\geq3-2=1.\]Equality holds when $a = b = c = \sqrt3).$
P6. (IneqMath) Let $x, y, z$ be positive real numbers such that $x+y+z=1$. Determine the largest constant $C$ such that the following inequality holds for all $x, y, z$:
\[(x+\frac{1}{x})(y+\frac{1}{y})(z+\frac{1}{z}) \geq C.\]S6. $C = \frac{1000}{27}$
Fixing $z$, we have $\left(x+\frac1x\right)\left(y+\frac1y\right) = xy+\frac1{xy}+(1-z) \ge 2+(1-z),$ where equality holds when (x=y).
Therefore, for any given ((x,y,z)), we can find
\[\left(\frac{x+y}{2},\frac{x+y}{2},z\right)\]for which the value of the given expression is smaller, and then find
\[\left(\frac{x+y}{2},\frac{\frac{x+y}{2}+z}{2},\frac{\frac{x+y}{2}+z}{2}\right)\]for which the value of the given expression is even smaller, and so on. By repeatedly taking the average of two variables in this way, we see that the minimum value of the given expression is attained when $x=y=z.$
P7. (IneqMath) Let $a, b, c > 0$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\left(\frac{a}{b+c}\right)^2 + \frac{b}{c+a} + \left(\frac{c}{a+b}\right)^2 \geq C.\]S7. $C = 1$
P8. (IneqMath) Let $a, b, c > 0$. Determine the largest constant $C$ such that the following inequality holds for all positive $a, b, c$:
\[\frac{c^2+a b}{a+b}+\frac{a^2+b c}{b+c}+\frac{b^2+c a}{c+a} \geq C(a+b+c).\]S8. $C = 1$
By rearrangement inequality,
P9. (IneqMath) Let $a, b, c > 0$ such that $a+b+c=2$ and $a^2+b^2+c^2=2$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraints:
\[a b c \leq C.\]S9. $C = \frac{4}{27}$
The support is the circle where the plane and the sphere are tangent, and $f(a,b,c)=abc$ decreases sharply when any one of the variables approaches 0.
Therefore, $abc$ is maximized when two of the three variables are equal and the remaining one is larger than them, in which case the solution is uniquely determined. The equality condition is that (a,b,c) are $\frac{4}{3}, \frac{1}{3}, \frac{1}{3}$, in any order.
P10. (IneqMath) Let $a, b, c$ be positive real numbers such that $a \geq b+c$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a^3+2 b^3+2 c^3}{a b c} \geq C.\]S10. $C = 6$
It seems like a simple AM-GM problem, but it’s an equal condition trick.
P11. (IneqMath) Let $a, b, c > 0$ and $k \in \mathbb{N}^{+}$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\left(a^2+\frac{2(k+1)^2}{b+k}\right)\left(b^2+\frac{2(k+1)^2}{c+k}\right)\left(c^2+\frac{2(k+1)^2}{a+k}\right) \geq (Ck+3)^3.\]S11. $C = 2$
By rearrangement inequality,
P12. (IneqMath) Let $a, b, c > 0$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{C}{(a+b+c)^2}\]S12. $C = \frac{27}{2}$
This rearrangement inequality uses the property that when the denominator approaches zero, the value increases sharply, so averaging the denominator makes the value smaller.
P13. (IneqMath) Let $a_1, a_2, a_3, \ldots, a_n$ be real numbers, where $n > 1$. Find the largest constant $C$ such that the following inequality holds for all $a_1, a_2, \ldots, a_n$:
\[\sqrt{a_1^2+\left(1-a_2\right)^2}+\sqrt{a_2^2+\left(1-a_3\right)^2}+\ldots +\sqrt{a_n^2+\left(1-a_1\right)^2} \geq Cn\]S13. $C = \frac{1}{\sqrt{2}}$
By rearrangement inequality,
P14. (IneqMath) Let $x$ and $y$ be two positive real numbers such that $x + y = 1$. Determine the largest constant $C$ such that the following inequality holds for all $x, y$:
\[\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq C.\]S14. $C = 9$
Since $f(s) = \frac{1}{s}$ increases sharply as $s$ approaches 0, it is important to prevent either value from becoming too small in order for $g(x, y) = \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right)$ to be minimized. Therefore, under the constraint $x + y = 1$, it is easy to see that the minimum occurs when $x = y = \frac{1}{2}$.
P15. (IneqMath) Let $a, b, c \geq 0$ such that $a^2+b^2+c^2+abc=4$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[a+b+c+\sqrt{\frac{a^2+b^2+c^2}{3}} \leq C\]S15. $C = 4$
Suppose one of the three variables is zero. Without loss of generality, let $c=0$. Then $a^2+b^2=4.$ Hence,
\[a+b+\sqrt{\frac{a^2+b^2}{3}} \le \sqrt{2a^2+2b^2}+\frac{2}{\sqrt{3}} =2\sqrt{2}+\frac{2}{\sqrt{3}} <4.\]Now consider the interior region, where $a,b,c>0$. The maximum can be found using the method of Lagrange multipliers. Let
\[\mathcal{L} = a+b+c+\sqrt{\frac{a^2+b^2+c^2}{3}}-\lambda(a^2+b^2+c^2+abc-4).\]Then
\[\frac{\partial \mathcal{L}}{\partial a} = 1+\frac{a}{\sqrt{3(a^2+b^2+c^2)}}-\lambda(2a+bc)=0.\]Subtracting $\frac{\partial \mathcal{L}}{\partial a}$ and $\frac{\partial \mathcal{L}}{\partial b}$ gives
\[\frac{a-b}{\sqrt{3(a^2+b^2+c^2)}} = \lambda(a-b)(2-c).\]If $a,b,c$ were all distinct, then
\[\lambda(2-a)=\lambda(2-b)=\lambda(2-c).\]Since $\lambda\neq 0$, this would imply $a=b=c$, which is a contradiction. Therefore, at least two variables must be equal. This is called UVW principle. Without loss of generality, we may assume $a=b.$ In this case, $c$ can be expressed as a function of $a$, so the desired inequality reduces to a one-variable problem. The maximum is then easily obtained. The equality condition is $a=b=c=1.$
P16. (IneqMath) In an acute triangle $ABC$, find the largest constant $C$ such that the following inequality holds for all angles $A, B, C$:
\[\frac{\cos A}{\cos B \cos C}+\frac{\cos B}{\cos C \cos A}+\frac{\cos C}{\cos A \cos B} \geq C\left(\frac{1}{1+\cos A}+\frac{1}{1+\cos B}+\frac{1}{1+\cos C}\right)\]S16. $C = 3$
Let us use the UVW principle. Set $x=\cos A,\qquad y=\cos B,\qquad z=\cos C.$ Then we know that $x^2+y^2+z^2+2xyz=1.$ Under a symmetric condition in three variables, when a symmetric inequality in three variables is handled by the method of Lagrange multipliers, the extremum is attained when at least two of the three variables are equal. Without loss of generality, let $x=y=t.$ Then from $x^2+y^2+z^2+2xyz=1,$ we get $z=1-2t^2.$ Thus the inequality reduces to a one-variable inequality in $t$.
P17. (IneqMath) Let $a, b, c$ be three non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \geq 0$:
\[\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right) \geq (ab+bc+ca-abc)^2 + C \cdot abc\]S17. $C = 4$
Solved by UVW principle.
P18. (IneqMath) Let $a, b, c$ be nonnegative real numbers such that $ab+bc+ca=4$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\left(a^2+b^2+c^2+1\right)\left(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}\right) \geq C.\]S18. $C = \frac{45}{8}$
Solved by UVW principle.
P19. (IneqMath) Let $a, b, c, m, n$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, m, n \in \mathbb{R}^{+}$:
\[\frac{a^2}{b(m a+n b)}+\frac{b^2}{c(m b+n c)}+\frac{c^2}{a(m c+n a)} \geq \frac{C}{m+n}.\]S19. $C = 3$
By rearrangement inequality,
The equality holds at $a = b = c$.
P20. (IneqMath) Given positive real numbers $a, b, c$ satisfying $ab + bc + ca = 3$, find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{1}{\sqrt{a+b}} + \frac{1}{\sqrt{c+b}} + \frac{1}{\sqrt{a+c}} \geq \sqrt{\frac{C}{a+b+c+3}}.\]S20. $C = 27$
Solved by UVW principle.
P21. (IneqMath) Given $a, b, c \geq 0$ satisfying $a b+b c+c a>0$, find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\sqrt{\frac{a}{b^2+b c+c^2}}+\sqrt{\frac{b}{c^2+c a+a^2}}+\sqrt{\frac{c}{a^2+a b+b^2}} \geq C \sqrt{\frac{a b+b c+c a}{(a+b)(b+c)(c+a)}}\]S21. $C = 2\sqrt{2}$
Solved by UVW principle.
P22. (IneqMath) Let $x, y, z \in [0,1]$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:
\[x(x-y)(z-x) \leq C.\]S22. $C = \frac{4}{27}$
It is sufficient to consider only the case $y < x < z$. For reference, the case $z < x < y$ is merely the same situation with the positions of $y$ and $z$ interchanged. For the left-hand side to become larger, $y$ should be smaller and $z$ should be larger, so let $y = 0$ and $z = 1$. Then,
\[\text{LHS} \le x^2(1-x) \le \frac{4}{27}\]with equality when $x = \frac{2}{3}$.
P23. (IneqMath) Let $a, b \geq 0$ and $(a+b)(1+4ab)=2$. Find the smallest constant $C$ such that the following inequality holds for all $a, b$ satisfying the given constraint:
\[ab(a+b+ab) \leq C\]S23. $C = \frac{5}{16}$
Let $p = a + b$ and $q = ab$. Then,
\[p(1 + 4q) = 2 \iff p = \frac{2}{1 + 4q}.\]Thus,
\[\text{LHS} = q(p + q) = q\left(\frac{2}{1 + 4q} + q\right).\]Since this is now a function of a single variable, its maximum value can be found easily.
P24. (IneqMath) Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a}{1+2 b^3}+\frac{b}{1+2 c^3}+\frac{c}{1+2 a^3} \geq C.\]S24. $C = 1$
Solved by UVW principle.
P25. (IneqMath) Let $a, b > 0$ and $a + b \leq \frac{3}{2}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b$ satisfying the given constraint:
\[\frac{a}{a^2+1} + \frac{4b}{b^2+4} \leq C.\]S25. $C = \frac{6}{5}$
Since
\[f(x)=\frac{4x}{x^2+4}\]attains its maximum at $x=2$, it is increasing on the interval $0<x<2$. Therefore, to maximize the left-hand side, we may take
\[b=\frac{3}{2}-a.\]Then the left-hand side becomes a function of a single variable, so its maximum value can be found easily.
P26. (IneqMath) Let $a, b, c, d \in \mathbb{R}^{+}$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$:
\[\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \geq \frac{C}{a+b+c+d}\]S26. $C = 64$
Titu’s lemma with equality at a
c:d = 1:1:2:4
P27. (IneqMath) Given non-negative real numbers $a, b, c$ satisfying $ab + bc + ca = 1$, find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{c}{\sqrt{a+b+3c}} + \frac{a}{\sqrt{c+b+3a}} + \frac{b}{\sqrt{a+c+3b}} \geq C.\]S27. $C = 1$
Solved by UVW principle.
P28. (IneqMath) Let $x, y$ be positive real numbers such that $x + y = 2$. Determine the smallest constant $C$ such that the following inequality holds for all $x, y$:
\[x^3 y^3 (x^3 + y^3) \leq C.\]S28. $C = 2$
Substituting $y = 2 - x$ into the left-hand side and differentiating with respect to $x$, we get
\[f(x) = x^3(2-x)^3\left(x^3 + (2-x)^3\right), f'(x) = -48x^2(2-x)^2(x-1)^3.\]Therefore, $f(x)$ attains its maximum at $x = 1$, so
\[f(x) \le f(1) = 2.\]P29. (IneqMath) Let $a, b, c, d \in \mathbb{R}$ such that $a^2 + b^2 + c^2 + d^2 = 4$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraint:
\[a^3 + b^3 + c^3 + d^3 \leq C.\]S29. $C = 8$
Even under the assumption that $a,b,c,d>0$, the above conclusion remains unchanged.
The given constraint is a circle, while the level curves of the given left-hand side are more convex than the circle.
Therefore, the maximum value of the left-hand side is attained when the absolute value of one of $a,b,c,d$ is equal to 2.
P30. (IneqMath) Find the largest constant $C$ such that for all real numbers $a$ and $b$, the following inequality holds:
\[\sqrt{a^2+b^2+\sqrt{2} a+\sqrt{2} b+1}+\sqrt{a^2+b^2-\sqrt{2} a-\sqrt{2} b+1} \geq C\]S30. $C = 2$
When $a^2 + b^2 = p$ is fixed, the left-hand side is maximized at $a+b=q=0$, and by symmetry, its value decreases as $q$ moves farther away from $0$. Therefore, the left-hand side is minimized when $q=\pm\sqrt{2p}.$ At this value,
\[\text{LHS}=\left\|\sqrt{p}+1\right\|+\left\|\sqrt{p}-1\right\|\ge 2,\]with equality when$p\le 1,\qquad q=\pm\sqrt{2p}.$
P31. (IneqMath) Let $a, b > 0$ such that $a + b = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b$:
\[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2 \geq C.\]S31. $C = \frac{25}{2}$
This rearrangement inequality uses the property that the square of a large number increases much more rapidly. The equality condition holds when $a = b = \frac{1}{2}$.
P32. (IneqMath) Given four real numbers $a, b, c, d \geq 0$ satisfying $a \geq b + 7c + d$, find the largest constant $C$ such that the following inequality holds for all $a, b, c, d$:
\[7(a+b)(b+c)(c+d)(d+a) \geq C a b c d\]S32. $C = 240$
First, by the rearrangement inequality, we have
\[7(a+b)(b+c)(c+d)(d+a) \ge 7\cdot 2a\cdot 2b\cdot 2c\cdot 2d =112abcd.\]This problem is equivalent to determining the coefficient $K$ of $abcd$ on the right-hand side under the additional constraint $a \ge b+7c+d.$ Using the rearrangement inequality trick, we observe that the left-hand side becomes smaller while the right-hand side being equal when the symmetric variables $b$ and $d$ are equal. Thus, it suffices to prove $7(a+b)(b+c)(c+b)(b+a) \ge K abcd$ under the constraint $a \ge 2b+7c.$ In general, when $a$ is sufficiently large, the left-hand side grows faster than the right-hand side as $a$ increases. However, when $a$ is only moderately large, the situation is not that simple. Now, if we gradually increase $K$ from 112, there will be a moment when a single point of tangency is formed at $a=2b+7c.$ At that moment, the critical value of $K$ is determined. This is called the KKT principle. Setting
\[a=2b+7c,\qquad x=\frac{b}{c},\]the problem becomes equivalent to proving
\[7(3x+7)^2(x+1)^2 \ge 240x^2(2x+7).\]Since this is now a single-variable inequality, its validity can be shown quite easily. For reference, the difference between the left-hand side and the right-hand side is tangent at $x=\frac{7}{3},$ which once again confirms the preceding KKT-based argument.
P33. (IneqMath) Let $x, y \in \mathbb{R}$ such that $(x+1)(y+2) = 8$. Find the largest constant $C$ such that the following inequality holds for all $x, y$ satisfying the given constraint:
\[\left\|xy - 10\right\| \geq C\]S33. $C = 8$
Expressing it as $y = f(x)$ makes the given inequality a single-variable inequality, so it can be solved easily.
P34. (IneqMath) Let $a, b, c$ be positive real numbers such that $abc = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq C.\]S34. $C = 3$
This rearrangement inequality uses the property that when the denominator approaches zero, the value increases rapidly, so averaging the denominator makes the value smaller. Therefore, the equality condition holds when $a = b = c = 1$.
P35. (IneqMath) Let $a, b, c > 0$. Find the largest constant $C$ such that the following inequality holds for all positive $a, b, c$:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{C \sqrt{a^2+b^2+c^2}}{\sqrt[3]{a b c}}\]S35. $C = \sqrt{3}$
Solved by UVW principle.
P36. (IneqMath) Let $x, y, z$ be positive real numbers such that $xy + yz + zx \geq 3$. Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint:
\[\frac{x}{\sqrt{4x+5y}} + \frac{y}{\sqrt{4y+5z}} + \frac{z}{\sqrt{4z+5x}} \geq C\]S36. $C = 1$
P37. (IneqMath) Given the real number $b \leq 2$, find the largest constant $C$ such that for all positive real numbers $x, y$ where $xy = 1$, the following inequality holds:
\[\frac{\sqrt{x^2+y^2}}{C}+\frac{b}{x+y} \geq 1+\frac{b}{2}\]S37. $C = \sqrt{2}$
By setting $x^2 + y^2 = t$, the left-hand side can be treated as a single-variable function, making it easier to solve.
P38. (IneqMath) Let $a, b, c > 0$. Find the smallest constant $C$ such that the following inequality holds for all positive $a, b, c$:
\[\sqrt{\frac{2 a b\left(a^2-a b+b^2\right)}{a^4+b^4}}+\sqrt{\frac{2 b c\left(b^2-b c+c^2\right)}{b^4+c^4}}+\sqrt{\frac{2 c a\left(c^2-c a+a^2\right)}{c^4+a^4}} \leq C \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{a+b+c}\]S38. $C = 1$
P39. (IneqMath) Let $a, b, c \in \mathbb{R}^{+}$ with $abc = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[a^3 + b^3 + c^3 + \frac{16}{(b+c)(c+a)(a+b)} \geq C.\]S39. $C = 5$
Solved by UVW principle.
P40. (IneqMath) Let $a > 1$ be a real number. Determine the largest constant $C$ such that for all $n \in \mathbb{N}$, the following inequality holds:
\[\frac{a^n-1}{n} \geq C\sqrt{a}^{n+1}-\sqrt{a}^{n-1}.\]S40. $C = 1$
P41. (IneqMath) Let $a, b, c \geq 0$ and $a+b+c=3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a+1}{a+b+1}+\frac{b+1}{b+c+1}+\frac{c+1}{c+a+1} \geq C.\]S41. $C = 2$
Solved by UVW principle.
P42. (IneqMath) Let $a, b, c$ be positive real numbers. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$:
\[\frac{a}{\sqrt{3 a+2 b+c}}+\frac{b}{\sqrt{3 b+2 c+a}}+\frac{c}{\sqrt{3 c+2 a+b}} \leq C \sqrt{a+b+c}\]S42. $C = \frac{1}{\sqrt{2}}$
P43. (IneqMath) Given $k > 1$ and $a, b, c > 0$ with $abc = k^3$, find the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\sum_{\text{cyc}}\left(\frac{1}{1+a}+\frac{1}{1+bc}\right) \geq C\left(\frac{1}{1+k}+\frac{1}{1+k^2}\right).\]S43. $C = 3
Solved by UVW principle.
P44. (IneqMath) Let $a, b, c$ be real numbers such that $a^2 + bc = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint:
\[\frac{1}{a^2+1} + \frac{1}{b^2+1} + \frac{1}{c^2+1} \leq C\]S44. $C = \frac{5}{2}$
By the UVW principle, the left-hand side is maximized when $b=c$. Let $a^2=t,\qquad b^2=c^2=1-t.$ Then the left-hand side becomes a single-variable function, so the problem can be solved easily. The equality condition is $a^2=1.$
P45. (IneqMath) Let $a, b, c \geq 0$ and $a^2 + b + c^2 = 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{1}{1+a^2} + \frac{1}{3(1+b)} + \frac{1}{1+c^2} \geq C.\]S45. $C = \frac{5}{3}$
Solved by UVW principle.
P46. (IneqMath) Let $x > 4$. Determine the largest constant $C$ such that the following inequality holds for all $x$:
\[\frac{x^3}{x-4} \geq C.\]S46. $C = 108$
The given inequality is a single-variable inequality, so it can be solved easily.
P47. (IneqMath) Let $a, b, c \geq 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[\sqrt{\frac{a b+b c+c a}{3}}-\sqrt[3]{a b c} \geq \frac{\sqrt{\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}}}{C}-\frac{1}{\sqrt[3]{a b c}}.\]S47. $C = \sqrt{3}$
Solved by UVW principle.
P48. (IneqMath) Let $a, b, c \geq 0$ with $ab+bc+ca > 0$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$:
\[\frac{a}{\sqrt{b+c+Ca}} + \frac{b}{\sqrt{a+c+Cb}} + \frac{c}{\sqrt{b+a+Cc}} \geq \sqrt{\frac{ab+bc+ca}{a+b+c}}\]S48. $C = 7$
By rearrangement inequality,
P49. (IneqMath) Let $x, y, z \in [-1, 3]$ such that $x + y + z = 3$. Find the smallest constant $C$ such that the following inequality holds for all $x, y, z$:
\[x^2 + y^2 + z^2 \leq C.\]S49. $C = 11$
It is easy to show that having at least one of $x, y, z$ positive gives a distance from the origin greater than or equal to the case where $x, y, z$ are all negative. Without loss of generality, assume $x \ge y \ge z$ and $x > 0$. If $z \ge 0$, then replacing $z$ by $-\left|z\right|$ is advantageous, since it allows the absolute values of $x$ and $y$ to increase. Since the square of a larger number gives a greater gain, by the rearrangement inequality, it is advantageous to make $z$ a larger negative number and move $x$ closer to 3. (Operation 1) If $x$ has not yet reached 3, then we should move $x$ closer to $3$ by exchanging values with $y$. (Operation 2) By repeatedly applying Operations 1 and 2, the maximum condition is attained at $x = 3$, $z = -1$, and $y = \pm 1$.
P50. (IneqMath) Let $p > q > 0$. Determine the largest constants $C$ such that the following inequality holds for all $x \in \mathbb{R}$:
\[\frac{p+q}{p-q} \geq C\frac{2(x^2-2 q x+p^2)}{x^2+2 q x+p^2}\]S50. $C = \frac{1}{2}$
You can expand the expression, and there is no equality condition.
Input: 2025.12.08 15:51