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
Input: 2025.12.08 15:51