For $a+b+c=2$ prove that $2a^ab^bc^c\geq a^2b+b^2c+c^2a$

Question

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 $a=b=c=\frac{2}{3}$.

Answer 1

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 $0<\varepsilon<c$ and $a+\varepsilon\geq c-\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 !

Answer 2

Proof: By taking logarithm on both sides, it suffices to prove that $$a\ln a + b\ln b + c\ln c \ge \ln \tfrac{a^2b + b^2c + c^2a}{2}.$$

We will use the following bounds (their proof is not hard and thus omitted): $$x\ln x \ge f(x) = \tfrac{2}{3}\ln \tfrac{2}{3} + (1 + \ln \tfrac{2}{3})(x- \tfrac{2}{3}) + \tfrac{9}{20}(x-\tfrac{2}{3})^2, \quad \forall x\in (0, 2]$$ and $$\ln\tfrac{4}{9} + \tfrac{9}{4}(y - \tfrac49) \ge \ln y, \quad \forall y > 0.$$

With the bounds above, it suffices to prove that $$f(a) + f(b) + f(c) \ge \ln\tfrac{4}{9} + \tfrac{9}{4}(\tfrac{a^2b + b^2c + c^2a}{2} - \tfrac49)$$ which is simplified to (by using $a+b+c=2$) $$18a^2+18b^2+18c^2+16 - 45a^2b - 45b^2c - 45c^2a \ge 0.$$ After homogenization, it suffices to prove that $$(18a^2+18b^2+18c^2)\tfrac{a+b+c}{2}+16(\tfrac{a+b+c}{2})^3- 45a^2b - 45b^2c - 45c^2a \ge 0.$$ The Buffalo Way works. WLOG, assume that $c = \min(a, b, c)$. Let $b = c + s, \ a = c + t$ for $s, t\ge 0$. It suffices to prove that $$(18s^2-18st+18t^2)c +11s^3+15s^2t-30st^2+11t^3\ge 0.$$ It is not hard to prove that $11s^3+15s^2t-30st^2+11t^3\ge 0$ for $s, t\ge 0$. We are done.

Answer 3

Remark: My first proof does not work for the stronger inequality below.

Let us prove a stronger inequality: $$2a^a b^b c^c \ge \frac{4}{27}(a + b + c)^3 - abc.$$ (Note: It is well-known that $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$.)

Letting $x = \frac32 a, y = \frac32 b, z = \frac32 c$, it suffices to prove that, for all $x, y, z > 0$ with $x + y + z = 3$, $$3 (x^x y^y z^z)^{2/3} \ge \frac{4}{27}(x+y+z)^3 - xyz. $$ (Note: We have $a^a b^b c^c = (2/3)^{2(x+y+z)/3}(x^xy^yz^x)^{2/3} = \frac49 (x^xy^yz^x)^{2/3} $.)

Using $\mathrm{e}^u \ge 1 + u$, we have $$(x^x y^y z^z)^{2/3} = \mathrm{e}^{\frac23(x\ln x + y\ln y + z\ln z)} \ge 1 + \frac23(x\ln x + y\ln y + z\ln z).$$

It suffices to prove that $$3 + 2x\ln x + 2y\ln y + 2z\ln z \ge \frac{4}{27}(x+y+z)^3 - xyz.$$

Fact 1: It holds that $u\ln u \ge u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3 = -\frac13 - \frac12 u + u^2 - \frac16 u^3$ for all $u > 0$.
(The proof is given at the end. Note: The RHS is $3$-th order Taylor approximation of $u\ln u$ around $u = 1$.)

Using Fact 1, it suffices to prove that \begin{align*} &3 + \left(-\frac23 - x + 2x^2 - \frac13 x^3\right) + \left(-\frac23 - y + 2y^2 - \frac13 y^3\right)\\ &\quad + \left(-\frac23 - z + 2z^2 - \frac13 z^3\right)\\ &\ge \frac{4}{27}(x+y+z)^3 - xyz \end{align*} or (using $x + y + z = 3$) $$3 - xy - yz - zx \ge 0$$ which is true.

We are done.

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Proof of Fact 1:

It suffices to prove that $$\ln u \ge \frac{1}{u}\left(u - 1 + \frac12(u-1)^2 - \frac16(u-1)^3\right).$$ Let $F(u) := \mathrm{LHS} - \mathrm{RHS}$. We have $$F'(u) = \frac{(u-1)^3}{3u^2}.$$ We have $F'(u) < 0$ on $(0, 1)$, and $F'(u) > 0$ on $(0, \infty)$, and $F'(1) = 0$. Thus, $F(u) \ge F(1) = 0$ for all $u > 0$.

We are done.