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I have inequality $$x\cdot\ln(x) + y\cdot\ln(y) \geq (x+y)\cdot \ln(x+y)$$ I transformated it to $$2^{x+y}\cdot x^x\cdot y^y\geq(x+y)^{x+y}$$ And I got stuck. Please help

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    As stated, the inequality does not hold. Plug $x=100$ and $y=200$ to check. – 2017-02-15
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    $x\ln x + y\ln y - (x+y)\ln (x+y)=x\ln \frac{x}{x+y} + y \ln\frac{y}{x+y}<0$ for $x,y>0$ – 2017-02-15
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    It's unclear where your $2^{x+y}$ comes from. It should be equivalent to $x^xy^y>(x+y)^{x+y}$. And, as others have noted, this is false (although the reverse inequality is true.) The true inequality is a statement about information theory, but that might be more than you wanted to know. :) – 2017-02-15
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    Ah, the $2^{x+y}$ comes because you probably wanted $x\ln x +y\ln y >(x+y)\ln\frac{x+y}{2}$, per Michael Rosenberg's answer. – 2017-02-15
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    I'm sorry for it. I already asked to delete the question – 2017-02-15

3 Answers 3

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It's obviously wrong! Try $x=y=1$.

By the way, since $f(x)=x\ln{x}$ is a convex function, we obtain: $$\frac{x\ln{x}+y\ln{y}}{2}\geq\frac{x+y}{2}\ln\frac{x+y}{2},$$ which gives $$x\ln{x}+y\ln{y}\geq(x+y)\ln\frac{x+y}{2}$$

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    Ah, that's probably the inequality OP wanted, since otherwise, it is hard to explain the $2^{x+y}$ is the work shown. – 2017-02-15
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Ok, it's false ...

But the map $f:(0,+\infty)\to\mathbb{R},t\mapsto x\ln(x)$ is convex and hence :

$$\forall(x,y)\in(0,+\infty)^2,\,f\left(\frac{x+y}{2}\right)\le\frac 12\left(f(x)+f(y)\right)$$

That is :

$$\forall(x,y)\in(0,+\infty)^2,\,x\ln(x)+y\ln(y)\ge(x+y)\ln\left(\frac{x+y}2\right)$$

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Contradictory example: Let $x=e=y$, the inequality claims $$ e+e > (2e)\cdot(1+\ln(2)) $$ which is wrong.