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For each $a,b \in \mathbb{R}$, show that: $$a^2 \log(a^2) + b^2 \log(b^2) - (a^2+b^2)\log(\frac{a^2+b^2}{2})\leq (a-b)^2$$

I've tried some things but nothing came out. I was thinking of using Taylor somehow but I can't see how to work it out.

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    Equivalently, if $f(x)=x\log(x)$, $\dfrac{f(x)+f(y)}{2}-f\left(\dfrac{x+y}{2}\right)\leq \dfrac{x+y}{2}-\sqrt{xy}$. – 2017-02-10

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Let $b=xa$, where $x\geq1$.

Hence, we need to prove that $f(x)\geq0$, where $$f(x)=(x-1)^2-2x^2\ln{x}+(x^2+1)\ln\frac{x^2+1}{2}.$$ But $$f'''(x)=\frac{4(x^2-1)}{x(x^2+1)^2}\geq0,$$ which gives $$f''(x)=2-4\ln{x}+2\ln\frac{x^2+1}{2}-\frac{4}{x^2+1}\geq0,$$ which gives $$f'(x)=2\ln\frac{x^2+1}{2}+2x-4\ln{x}-2\geq0$$ and $$f(x)\geq f(1)=0.$$ Done!

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    Do you mean $b=xa$ in the first line? – 2017-02-10
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    @Martin R I fixed my post. Thank you! – 2017-02-10