Inequality Proof Problems [351-400]
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Restructured the IneqMath training data.
P351. Let $a, b \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$:
\[C(a^3 + b^3) \geq (a + b)^3\]S351. $C = 4$
The function \(f(x)=x^{3}\) is convex on \((0,+\infty)\), thus by Jensen’s inequality,
\[\left(\frac{a+b}{2}\right)^{3}\le \frac{a^{3}+b^{3}}{2}.\]Multiplying both sides by \(8\) gives
\[(a+b)^{3}\le 4(a^{3}+b^{3}),\]i.e.
\[4(a^{3}+b^{3})\ge (a+b)^{3}.\]Equality holds when \(a=b\), since then
\[4(a^{3}+b^{3})=4(2a^{3})=8a^{3}=(2a)^{3}=(a+b)^{3}.\]Therefore the sharp constant is
\[C=4.\]P352. Let $x, y, z > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $x, y, z$:
\[(x y + y z + z x) \left( \frac{1}{(x + y)^2} + \frac{1}{(y + z)^2} + \frac{1}{(z + x)^2} \right) \geq C.\]S352. $C = \frac{9}{4}$
The given inequality
\[(xy+yz+zx)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right)\ge \frac{9}{4}\]is equivalent (after multiplying by \((x+y)^{2}(y+z)^{2}(z+x)^{2}\)) to
\[4(xy+yz+zx)\left((z+x)^{2}(y+z)^{2}+(x+y)^{2}(z+x)^{2}+(x+y)^{2}(y+z)^{2}\right) \ge 9(x+y)^{2}(y+z)^{2}(z+x)^{2}. \qquad (1)\]Let us denote \(p=x+y+z\), \(q=xy+yz+zx\), \(r=xyz\). Using the standard identities
\[(x+y)(y+z)(z+x)=pq-r,\]hence
\[(x+y)^{2}(y+z)^{2}(z+x)^{2}=(pq-r)^{2},\]and
\[(x+y)^{2}(y+z)^{2}+(y+z)^{2}(z+x)^{2}+(z+x)^{2}(x+y)^{2} =(p^{2}+q)^{2}-4p(pq-r),\]we can rewrite (1) as
\[4q\left((p^{2}+q)^{2}-4p(pq-r)\right)\ge 9(pq-r)^{2}.\]Expanding and simplifying gives
\[4p^{4}q-17p^{2}q^{2}+4q^{3}+34pqr-9r^{2}\ge 0,\]or equivalently
\[3pq\left(p^{3}-4pq+9r\right) +q\left(p^{4}-5p^{2}q+4q^{2}+6pr\right) +r(pq-9r)\ge 0.\]This holds by the known nonnegative symmetric inequalities \(N_{1}\), \(N_{2}\), \(N_{3}\) together with \(p,q,r>0\). Hence (1) is true.
Equality holds if and only if \(x=y=z\). Indeed, when \(x=y=z\) we have
\[xy+yz+zx=3x^{2},\]and
\[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}} =\frac{3}{(2x)^{2}} =\frac{3}{4x^{2}}.\]Thus the left-hand side equals
\[3x^{2}\cdot \frac{3}{4x^{2}}=\frac{9}{4}.\]So the minimum value is \(\frac{9}{4}\) and the sharp constant is
\[C=\frac{9}{4}.\]Therefore, the answer is \(C=\frac{9}{4}\).
P353. Let $a, b, c$ be the lengths of the sides of a triangle. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$:
\[(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq C (a^{a} b^{b} c^{c}).\]S353. $C = 1$
By the weighted power mean inequality we have
\[\sqrt[a+b+c]{\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c} } \le \frac{1}{a+b+c}\left(a\cdot\frac{a+b-c}{a}+b\cdot\frac{b+c-a}{b}+c\cdot\frac{c+a-b}{c}\right).\]But
\[a\cdot\frac{a+b-c}{a}+b\cdot\frac{b+c-a}{b}+c\cdot\frac{c+a-b}{c} =(a+b-c)+(b+c-a)+(c+a-b)=a+b+c,\]so the right-hand side equals \(1\). Hence
\[\sqrt[a+b+c]{\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c}} \le 1.\]Raising both sides to the power \(a+b+c\) gives
\[\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c}\le 1,\]i.e.
\[(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c}\le a^{a}b^{b}c^{c}.\]Equality holds when
\[\frac{a+b-c}{a}=\frac{b+c-a}{b}=\frac{c+a-b}{c},\]which implies \(a=b=c\) (an equilateral triangle). Therefore the sharp constant is \(C=1.\)
P354. Let $x, y, z \in \mathbb{R}^{+}$ such that $x y z = 1$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq x + y + z$. Determine the largest constant $C$ such that for any natural number $n$, the following inequality holds for all $x, y, z$:
\[\frac{1}{x^{n}} + \frac{1}{y^{n}} + \frac{1}{z^{n}} \geq C (x^{n} + y^{n} + z^{n}).\]S354. $C = 1$
After setting
\[x=\frac{a}{b},\quad y=\frac{b}{c},\quad z=\frac{c}{a},\]we have
\[xyz=1.\]The initial condition
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge x+y+z\]becomes
\[\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \iff a^{2}b+b^{2}c+c^{2}a\ge ab^{2}+bc^{2}+ca^{2} \iff (a-b)(b-c)(c-a)\le 0.\]Let \(n\in\mathbb{N}\) and take
\[A=a^{n},\quad B=b^{n},\quad C=c^{n}.\]Then the order of \(A,B,C\) is the same as the order of \(a,b,c\), so
\[(a-b)(b-c)(c-a)\le 0 \iff (A-B)(B-C)(C-A)\le 0.\]Applying the equivalence above to \(A,B,C\) yields
\[\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\ge \frac{A}{B}+\frac{B}{C}+\frac{C}{A}.\]But
\[\frac{B}{A}=\left(\frac{b}{a}\right)^{n}=\frac{1}{x^{n}},\quad \frac{C}{B}=\left(\frac{c}{b}\right)^{n}=\frac{1}{y^{n}},\quad \frac{A}{C}=\left(\frac{a}{c}\right)^{n}=\frac{1}{z^{n}},\]and
\[\frac{A}{B}=x^{n},\quad \frac{B}{C}=y^{n},\quad \frac{C}{A}=z^{n}.\]Hence we obtain
\[\frac{1}{x^{n}}+\frac{1}{y^{n}}+\frac{1}{z^{n}}\ge x^{n}+y^{n}+z^{n}.\]Therefore the largest constant \(C\) such that
\[\frac{1}{x^{n}}+\frac{1}{y^{n}}+\frac{1}{z^{n}}\ge C\left(x^{n}+y^{n}+z^{n}\right)\]always holds is
\[C=1.\]Equality holds when \(a=b=c\), which corresponds to \(x=y=z=1\), and then
\[\frac{1}{x^{n}}+\frac{1}{y^{n}}+\frac{1}{z^{n}}=3 \quad\text{and}\quad x^{n}+y^{n}+z^{n}=3,\]so the inequality becomes \(3\ge C\cdot 3\), i.e. \(C=1\).
P355. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$:
\[\frac{1}{1-x y}+\frac{1}{1-y z}+\frac{1}{1-z x} \leq C.\]S355. $C = \frac{27}{8}$
Let
\[p=x+y+z=1,\quad q=xy+yz+zx,\quad r=xyz.\]It is easy to check that
\[(1-xy)(1-yz)(1-zx)=1-q+pr-r^{2},\]and
\[(1-xy)(1-yz)+(1-yz)(1-zx)+(1-zx)(1-xy)=3-2q+pr.\]Hence the given inequality
\[\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\ge \frac{27}{8}\]is equivalent (by Cauchy-Schwarz) to
\[\left(\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\right) \left((1-xy)+(1-yz)+(1-zx)\right)\ge 9,\]i.e.
\[\left(\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\right) (3-q)\ge 9.\]So it is enough to prove the sharper statement
\[\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\ge \frac{9}{3-q}. \qquad (1)\]Now apply Cauchy-Schwarz again in the form
\[\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx} \ge \frac{(1+1+1)^{2}}{(1-xy)+(1-yz)+(1-zx)} =\frac{9}{3-q},\]which proves (1).
It remains to show
\[\frac{9}{3-q}\ge \frac{27}{8} \quad\Leftrightarrow\quad 3-q\le \frac{8}{3} \quad\Leftrightarrow\quad q\ge \frac{1}{3}.\]But since \(x+y+z=1\),
\[q=xy+yz+zx\le \frac{(x+y+z)^{2}}{3}=\frac{1}{3},\]and in fact
\[q\le \frac{1}{3},\]so
\[\frac{9}{3-q}\ge \frac{9}{3-\frac{1}{3}}=\frac{27}{8}.\]Therefore,
\[\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\ge \frac{27}{8}.\]Equality holds when \(x=y=z=\frac{1}{3}\). Indeed, then
\[xy=yz=zx=\frac{1}{9},\]so
\[\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx} =3\cdot \frac{1}{1-\frac{1}{9}} =3\cdot \frac{9}{8} =\frac{27}{8}.\]Thus the sharp constant is
\[C=\frac{27}{8}.\]P356. Let $\alpha_{i}>0, i=1,2, \ldots, n$, be real numbers such that $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=1$. Determine the maximal constant $C$ such that the following inequality holds for all $\alpha_{i}$:
\[\alpha_{1}^{\alpha_{1}} \alpha_{2}^{\alpha_{2}} \cdots \alpha_{n}^{\alpha_{n}} \geq C.\]S356. $C = \frac{1}{n}$
If we take
\[a_i=\frac{1}{\alpha_i},\quad i=1,2,\ldots,n,\]then by the weighted \(AM\ge GM\) inequality (with weights \(\alpha_1,\alpha_2,\ldots,\alpha_n\) and \(\alpha_1+\alpha_2+\cdots+\alpha_n=1\)) we get
\[\prod_{i=1}^{n} a_i^{\alpha_i} \le \sum_{i=1}^{n}\alpha_i a_i.\]Substituting \(a_i=\frac{1}{\alpha_i}\) gives
\[\prod_{i=1}^{n}\left(\frac{1}{\alpha_i}\right)^{\alpha_i} \le \sum_{i=1}^{n}\alpha_i\cdot \frac{1}{\alpha_i} =\sum_{i=1}^{n}1 =n.\]Thus
\[\frac{1}{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\cdots \alpha_n^{\alpha_n}}\le n,\]i.e.
\[\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\cdots \alpha_n^{\alpha_n}\ge \frac{1}{n}.\]Equality holds when all \(a_i\) are equal, i.e.
\[\frac{1}{\alpha_1}=\frac{1}{\alpha_2}=\cdots=\frac{1}{\alpha_n},\]which means
\[\alpha_1=\alpha_2=\cdots=\alpha_n=\frac{1}{n}.\]Then
\[\alpha_1^{\alpha_1}\cdots \alpha_n^{\alpha_n} =\left(\frac{1}{n}\right)^{\alpha_1+\cdots+\alpha_n} =\left(\frac{1}{n}\right)^{1} =\frac{1}{n}.\]Therefore, the sharp constant is
\[C=\frac{1}{n}.\]P357. Let $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$. Find the smallest constant $C$ such that the following inequality holds for all $a, b \in \mathbb{R}^{+}$ and $n \in \mathbb{N}$:
\[(a+b)^{n}\left(a^{n}+b^{n}\right) \leq C\left(a^{2 n}+b^{2 n}\right)\]S357. $C = 2^n$
By the power mean inequality, for any \(x,y\in\mathbb{R}^{+}\) and \(n\in\mathbb{N}\), we have
\[\left(\frac{x+y}{2}\right)^{n}\le \frac{x^{n}+y^{n}}{2}.\]Therefore,
\[\begin{aligned} (a+b)^{n}(a^{n}+b^{n}) &=2^{n}\left(\frac{a+b}{2}\right)^{n}(a^{n}+b^{n}) \\ &\le 2^{n}\left(\frac{a^{n}+b^{n}}{2}\right)(a^{n}+b^{n}) =2^{n-1}(a^{n}+b^{n})^{2}. \end{aligned}\]Also,
\[(a^{n}+b^{n})^{2}\le 2(a^{2n}+b^{2n}),\]which follows from \((u-v)^{2}\ge 0\) with \(u=a^{n}\) and \(v=b^{n}\).
Hence,
\[2^{n-1}(a^{n}+b^{n})^{2}\le 2^{n-1}\cdot 2(a^{2n}+b^{2n})=2^{n}(a^{2n}+b^{2n}).\]Combining the two estimates gives
\[(a+b)^{n}(a^{n}+b^{n})\le 2^{n}(a^{2n}+b^{2n}).\]Equality holds when \(a=b\), since then
\[(a+b)^{n}(a^{n}+b^{n})=(2a)^{n}\cdot 2a^{n}=2^{n+1}a^{2n}\]and
\[2^{n}(a^{2n}+b^{2n})=2^{n}\cdot 2a^{2n}=2^{n+1}a^{2n}.\]Therefore, the sharp constant is
\[C=2^{n}.\]P358. Let $a_{1}, a_{2}, \ldots, a_{n} \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$:
\[\sum_{k=1}^{n} k a_{k} \leq C + \sum_{k=1}^{n} a_{k}^{k}\]S358. $C = \binom{n}{2}$
For \(1\le k\le n\) we have
\[a_k^{k}+(k-1)=a_k^{k}+\underbrace{1+1+\cdots+1}_{k-1} \ge k\sqrt[k]{a_k^{k}\cdot \underbrace{1\cdot 1\cdots 1}_{k-1}} =ka_k,\]by \(AM\ge GM\).
After adding these inequalities for \(k=1,2,\ldots,n\) we get
\[\sum_{k=1}^{n}ka_k \le \sum_{k=1}^{n}\left(a_k^{k}+k-1\right) =\sum_{k=1}^{n}a_k^{k}+\sum_{k=1}^{n}(k-1) =\sum_{k=1}^{n}a_k^{k}+\frac{n(n-1)}{2} =\sum_{k=1}^{n}a_k^{k}+\binom{n}{2}.\]Equality holds when equality holds in each \(AM\ge GM\) step, i.e. when
\[a_k^{k}=1 \quad\Rightarrow\quad a_k=1\]for all \(k\). Then
\[\sum_{k=1}^{n}ka_k=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\]and
\[\sum_{k=1}^{n}a_k^{k}+\binom{n}{2}=n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2},\]so equality indeed holds.
Therefore, the sharp constant is
\[C=\binom{n}{2}.\]P359. Let $a, b, c$ be positive numbers. Consider the following inequality:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \quad () \quad \frac{5 a}{2 a^2+b^2+c^2}+\frac{5 b}{a^2+2 b^2+c^2}+\frac{5 c}{a^2+b^2+2 c^2}.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S359. (F) None of the above
P360. Let $a, b, c$ be positive real numbers such that $4abc = a + b + c + 1$. Consider the following inequality:
\[\frac{b^2+c^2}{2a}+\frac{c^2+a^2}{3b}+\frac{b^2+a^2}{c} \quad () \quad 2(ab+bc+ca).\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S360. (F) None of the above
P361. Let $a, b, c > 0$ be positive real numbers such that $a+b+c \geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$. Consider the following inequality:
\[\frac{a}{2a+1}+\frac{b}{b+1}+\frac{c}{c+1} \quad () \quad 2.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S361. (F) None of the above
P362. Let $a, b, c$ be positive real numbers such that $abc = 1$. Consider the following inequality:
\[\sum \frac{1}{b(a+b)} \quad () \quad \frac{5}{3}.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S362. (F) None of the above
P363. Let $a, b, c$ be positive real numbers such that $abc = 1$. Consider the following inequality:
\[\sum \frac{a(a^3+1)}{3b+c} \quad () \quad 2a^3 + b^3 + c^3.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S363. (F) None of the above
P364. Let $a, b, c$ be non-negative numbers such that $ab + ac + bc \neq 0$. Consider the following inequality:
\[a^3 + b^3 + c^3 + 2abc + \frac{4a^2b^2c^2}{a^3 + b^3 + c^3 + abc} \quad () \quad \sum_{\text{cyc}}(2a^2b + 3a^2c) .\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S364. (F) None of the above
P365. Let $a, b, c \in \mathbb{R}^{+}$. Consider the following inequality:
\[\sqrt{\frac{a+b}{2c}} + \sqrt{\frac{b+c}{2a}} + \sqrt{\frac{c+a}{b}} \quad () \quad \sqrt{\frac{8(a+b+c)}{\sqrt[3]{abc}}}.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S365. (F) None of the above
P366. Let $a, b, c \geq 0$ be non-negative real numbers such that $a + b + c = 1$. Consider the following inequality:
\[a^3 + b^3 + c^3 + 6abc \quad () \quad \frac{1}{3}.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S366. (F) None of the above
P367. Let $a, b, c \neq 0$ be non-zero real numbers such that $a^2 + b^2 + c^2 = 2(ab + bc + ca)$. Consider the following expression:
\[(a+b+c)\left(\frac{1}{2a}+\frac{1}{3b}+\frac{1}{c}\right) \quad () \quad 14.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S367. (F) None of the above
P368. Let $a, b, c \geq 0$ be non-negative real numbers such that $a^2 + b + c^2 = 1$. Consider the following inequality:
\[\frac{1}{1+a^2} + \frac{1}{3(1+b)} + \frac{1}{1+c^2} \quad () \quad \frac{7}{4}.\]Determine the correct inequality relation to fill in the blank.
Options:
(A) $\leq$
(B) $\geq$
(C) $=$
(D) $<$
(E) $>$
(F) None of the above
S368. (F) None of the above
P323. What is the minimum value of the following expression?
S323.
Rather than solving the problem by brute force, we should solve it using a trigonometric substitution.
Therefore, the minimum value of $y = f(x)$ is $-\frac{1}{2}.$ We must also check whether the minimum value is actually feasible. Since $A+B=\arctan x+\arctan e^x$ can take all possible values, the result is valid. For reference, the actual shape of the function ( y=f(x) ) is as follows.
Input: 2025.12.08 15:51