umedkarimov wrote: Problem 4. Let a,b be positive real numbers and let x,y be positive real numbers less than 1, such that: a/(1-x)+b/(1-y)=1 Prove that: ∛ay+∛bx≤1. Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[3]{ay}+\sqrt[3]{bx}\leq 1.$$

$3=\frac{a}{1-x}+(1-x)+y+\frac{b}{1-y}+(1-y)+x \geq 3(\sqrt[3]{ay}+\sqrt[3]{bx})$

Or by Holder. \[ \left( \frac{a}{1-x}+\frac{b}{1-y} \right) (1-x + 1-y)(y + x) \geq \left(\sqrt[3]{ay}+\sqrt[3]{bx}\right)^3 \]and we are left with \[ 1 \geq (2- x-y)(x+y) \Rightarrow (x+y - 1)^2 \geq 0\]Done.

RagvaloD wrote: $3=\frac{a}{1-x}+(1-x)+y+\frac{b}{1-y}+(1-y)+x \geq 3(\sqrt[3]{ay}+\sqrt[3]{bx})$ Equality holds when $a=b=\frac{1}{4} , x=y=\frac{1}{2}.$ Very nice.

Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}.$$here

sqing wrote: Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}.$$here Solution. From the inequality AM-GM $4\sqrt[4]{u_1u_2u_3u_4}\le u_1+u_2+u_3+u_4$ it follows that \begin{align*}\frac{a}{1-x}+\frac{b}{1-y}=1\Longrightarrow&\left(\frac{1}{2}+\frac{a}{1-x}+(1-x)+y\right)+\left(x+(1-y)+\frac{b}{1-y}+\frac{1}{2}\right)=4\\ \Longrightarrow&4\sqrt[4]{\frac{ay}{2}}+4\sqrt[4]{\frac{bx}{2}}\le4\\ \Longrightarrow&\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}, \end{align*}where the equality occurs when $a=b=\frac{1}{4}$ and $x=y=\frac{1}{2}$. As desired. $\blacksquare$

ytChen wrote: sqing wrote: Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}.$$here Solution. From the inequality AM-GM $4\sqrt[4]{u_1u_2u_3u_4}\le u_1+u_2+u_3+u_4$ it follows that \begin{align*}\frac{a}{1-x}+\frac{b}{1-y}=1\Longrightarrow&\left(\frac{1}{2}+\frac{a}{1-x}+(1-x)+y\right)+\left(x+(1-y)+\frac{b}{1-y}+\frac{1}{2}\right)=4\\ \Longrightarrow&4\sqrt[4]{\frac{ay}{2}}+4\sqrt[4]{\frac{bx}{2}}\le4\\ \Longrightarrow&\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}, \end{align*}where the equality occurs when $a=b=\frac{1}{4}$ and $x=y=\frac{1}{2}$. As desired. $\blacksquare$ Yeah. Thanks. Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ For the integer $n\geq 3$ ,prove that $$\sqrt[n]{ay}+\sqrt[n]{bx}\leq \sqrt[n]{2^{n-3}}.$$

sqing wrote: ytChen wrote: sqing wrote: Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}.$$here Solution. From the inequality AM-GM $4\sqrt[4]{u_1u_2u_3u_4}\le u_1+u_2+u_3+u_4$ it follows that \begin{align*}\frac{a}{1-x}+\frac{b}{1-y}=1\Longrightarrow&\left(\frac{1}{2}+\frac{a}{1-x}+(1-x)+y\right)+\left(x+(1-y)+\frac{b}{1-y}+\frac{1}{2}\right)=4\\ \Longrightarrow&4\sqrt[4]{\frac{ay}{2}}+4\sqrt[4]{\frac{bx}{2}}\le4\\ \Longrightarrow&\sqrt[4]{ay}+\sqrt[4]{bx}\leq \sqrt[4]{2}, \end{align*}where the equality occurs when $a=b=\frac{1}{4}$ and $x=y=\frac{1}{2}$. As desired. $\blacksquare$ Yeah. Thanks. Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ For the integer $n\geq 3$ ,prove that $$\sqrt[n]{ay}+\sqrt[n]{bx}\leq \sqrt[n]{2^{n-3}}.$$ Solution. From the inequality AM-GM it follows that \begin{align*}\frac{a}{1-x}+\frac{b}{1-y}=1\Longrightarrow&\left(\frac{n-3}{2}+\frac{a}{1-x}+(1-x)+y\right)+\left(x+(1-y)+\frac{b}{1-y}+\frac{n-3}{2}\right)=n\\ \Longrightarrow&n\sqrt[n]{\frac{ay}{2^{n-3}}}+n\sqrt[n]{\frac{bx}{2^{n-3}}}\le n\\ \Longrightarrow&\sqrt[n]{ay}+\sqrt[n]{bx}\leq \sqrt[n]{2^{n-3}}, \end{align*}where the equality occurs when $a=b=\frac{1}{4}$ and $x=y=\frac{1}{2}$. As desired. $\blacksquare$

ytChen wrote: sqing wrote: Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ For the integer $n\geq 3$ ,prove that $$\sqrt[n]{ay}+\sqrt[n]{bx}\leq \sqrt[n]{2^{n-3}}.$$ Solution. From the inequality AM-GM it follows that \begin{align*}\frac{a}{1-x}+\frac{b}{1-y}=1\Longrightarrow&\left(\frac{n-3}{2}+\frac{a}{1-x}+(1-x)+y\right)+\left(x+(1-y)+\frac{b}{1-y}+\frac{n-3}{2}\right)=n\\ \Longrightarrow&n\sqrt[n]{\frac{ay}{2^{n-3}}}+n\sqrt[n]{\frac{bx}{2^{n-3}}}\le n\\ \Longrightarrow&\sqrt[n]{ay}+\sqrt[n]{bx}\leq \sqrt[n]{2^{n-3}}, \end{align*}where the equality occurs when $a=b=\frac{1}{4}$ and $x=y=\frac{1}{2}$. As desired. $\blacksquare$ Thanks.

sqing wrote:

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Let $a,b$ be positive real numbers and let $ x,y$ be positive real numbers less than 1, such that $ \frac{a}{1-x}+\frac{b}{1-y}=1.$ Prove that $$\sqrt[3]{ay}+\sqrt[3]{bx}\leq 1.$$ $$2-x-y=(2-x-y)(\frac{a}{1-x}+\frac{b}{1-y})\geq(\sqrt{a}+\sqrt{b})^2\geq 2(\sqrt{a}+\sqrt{b})-1$$$$\sqrt[3]{ay}+\sqrt[3]{bx}\leq\frac{1}{3} (2\sqrt{a}+x+2\sqrt{b}+y)\leq1.$$

The $\sqrt[3]{}$ hints for an AM-GM with three numbers and the given condition hints to use $\frac{a}{1-x}$ and $\frac{b}{1-y}$. We have $\sqrt[3]{ay} = \sqrt[3]{\frac{a}{1-x} \cdot (1-x) \cdot y} \leq \frac{1}{3}\left(\frac{a}{1-x} + 1-x + y\right)$; similarly $\sqrt[3]{bx} \leq \frac{1}{3}\left(\frac{b}{1-y} + 1-y + x\right)$, thus $\sqrt[3]{ay} + \sqrt[3]{bx} \leq \frac{1}{3}\left(\frac{a}{1-x} + \frac{b}{1-y}\right) + \frac{1}{3} \cdot 2 = 1$. As a bonus, note that equality holds (due to the AM-GM applications) if and only if $a = (1-x)^2 = y^2$, $b = (1-y)^2 = x^2$ and $x+y = 1$.