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Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove that: $$2a^ab^bc^c\geq a^2b+b^2c+c^2a$$ I tried convexity, but without success:

We need to prove that $$\ln2+\sum_{cyc}a\ln{a}\geq\ln\sum_{cyc}a^2b$$ and since $f(x)=x\ln{x}$ is a convex function, by Jensen we obtain: $$\ln2+\sum_{cyc}a\ln{a}\geq\ln2+3\cdot\frac{2}{3}\ln\frac{2}{3}=\ln\frac{8}{9}.$$ Thus, we need to prove that $$\frac{8}{9}\geq\sum_{cyc}a^2b$$ or $$(a+b+c)^3\geq9(a^2b+b^2c+c^2a),$$ which is wrong for $c\rightarrow0^+$.

The equality occurs for $a=b=c=\frac{2}{3}$.

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    $\sqrt[a+b+c]{a^ab^bc^c}$ looks like some kind of weighted geometric mean, although I'm a bit unsure about exactly how it behaves. – 2017-01-07
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    @Arthur If so, what we'll do with $2$? By the way it can be $\leq$. – 2017-01-07
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    Expand 2 as $a+b+c$, then we get $$a^{a+1}b^bc^c+b^{b+1}a^ac^c+c^{c+1}b^ba^a \geq a^2b+b^2c+c^2a$$ I think this can be proven with weighted AM-GM, but I'm not sure. Maybe dividing by $abc$ helps, then we get $$a^{a}b^{b-1}c^{c-1}+b^{b}a^{a-1}c^{c-1}+c^{c}b^{b-1}a^{a-1} \geq \frac{a}c+\frac{b}{a}+\frac{c}{b}$$ – 2017-01-07
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    If the minimum to $f=2a^ab^bc^c-a^2b-b^2a-c^2a$ is at a unique point then the inequality follows https://www.ma.utexas.edu/users/ntran/symmetry.pdf. – 2017-01-07
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    This inequality can be proven using the same strategy as used here: http://math.stackexchange.com/a/1852608/393239 – 2017-04-04
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    i have a very simple proof of it do you want hint or a complete proof ? – 2018-07-27
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    I think it's better show a full solution. – 2018-07-27

1 Answers 1

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Hints :In fact the key of this inequality is the following lemma :

$$2(a+\varepsilon)^{a+\varepsilon}b^b(c-\varepsilon)^{c-\varepsilon}\geq 2a^a b^b c^c \geq a^2b+b^2c+c^2a\geq (a+\varepsilon)^2b+b^2(c-\varepsilon)+(a+\varepsilon)(c-\varepsilon)^2$$ With $0<\varepsilon

Proof : We have to prove (for the first part) $$2(a+\varepsilon)^{a+\varepsilon}b^b(c-\varepsilon)^{c-\varepsilon}\geq 2a^a b^b c^c$$ Wich is equivalent to :

$$(a+\varepsilon)ln(a+\varepsilon)+(c-\varepsilon)ln(c-\varepsilon)\geq aln(a)+cln(c)$$

Now we use Niculescu's inequality to solve this :

Let $a,b,c,d$ be real positive numbers such that$$ a\geq c , b\leq d ,c\geq d $$ And $f$ be a convex function we have :

$$0.5(f(a)+f(b))-f((a+b)0.5)\geq 0.5(f(c)+f(d))-f(0.5(c+d))$$

It's easy to remark that we have :

$$ a+\varepsilon\geq a , c\geq c-\varepsilon ,a+\varepsilon\geq c-\varepsilon $$ And that $xln(x)$ is convex . The second part is easy to show and now you can build the inequality !