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Hello I'm interested by the following problem :

Let $x,y>0$ then we have : $$2\sqrt{(x^xy^y)^{\frac{1}{x+y}}\sqrt{xy}}\leq x+y$$

My try : Since the inequality is homogeneous we can put $x=1$ it gives : $$2\sqrt{(y^y)^{\frac{1}{1+y}}\sqrt{y}}\leq 1+y$$

Wich is true graphically

My question is how to prove the one variable inequality without using the derivative .

Many thanks (iff you have a hint obviously ).

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    $\begingroup$ Are you sure about the 2 in the LHS? With $x=y$ the LHS becomes $2x$ but the RHS is $x$. $\endgroup$ – user10354138 Jun 22 at 15:20
  • $\begingroup$ What makes you think that is homogeneous? Just another random question and statements afterwards? $\endgroup$ – Macavity Jun 22 at 15:31
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I think it means that we need to prove that: $$\sqrt{(x^xy^y)^{\frac{1}{x+y}}\sqrt{xy}}\leq \frac{x+y}{2},$$ otherwise, your inequality is wrong for $x=y=1$. Let $y=kx$.

Since our inequality is symmetric, we can assume $k\geq1.$

Thus, we need to prove that $$x^xy^y\leq \left(\frac{(x+y)^2}{4\sqrt{xy}}\right)^{x+y}$$ or $$x^x(kx)^{kx}\leq\left(\frac{x(1+k)^2}{4\sqrt{k}}\right)^{x(1+k)}$$ or $$x(kx)^k\leq\left(\frac{x(1+k)^2}{4\sqrt{k}}\right)^{1+k}$$ or $$k^k\leq \left(\frac{(1+k)^2}{4\sqrt{k}}\right)^{1+k}$$ or $f(k)\geq0,$ where $$f(k)=2(1+k)\ln(1+k)-k\ln{k}-\frac{1+k}{2}\ln{k}-(k+1)\ln4.$$ Now, we obtain: $$f'(k)=2\ln(1+k)+2-\ln{k}-1-\frac{1}{2}\ln{k}-\frac{1+k}{2k}-\ln4,$$ $$f''(k)=\frac{2}{1+k}-\frac{3}{2k}+\frac{1}{2k^2}=\frac{(k-1)^2}{2k^2(1+k)}\geq0,$$ which gives $$f'(k)\geq f(1)=0$$ and from here $$f(k)\geq f(1)=0$$ and we are done!

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The inequality is wrong, try x=y=1.

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Use $(y^y)^{1/y} = y$ and that you take the $1/(y+1)$th root.

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