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Let $a$ and $b$ be positive numbers such that $a+b=2$. Prove that: $$a^a+b^b+3\sqrt[3]{a^2b^2}\geq5$$

My trying.

Easy to show that $x^x\geq\frac{x^3-x^2+x+1}{2}$ for all $x>0$, but $$\frac{a^3-a^2+a+1}{2}+\frac{b^3-b^2+b+1}{2}+3\sqrt[3]{a^2b^2}\geq5$$ is wrong for $a\rightarrow0^+$

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    What happens if you plug in $b=2-a$ and just differentiate the LHS and find its min using single variable calc? I don't feel like doing the calculations but it seems like it should work – 2017-02-25
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    @mathworker21 I tried. I think it's nothing. – 2017-02-25
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    I tried to solve the derivative=0 on wolfram alpha, and it exceeded computation time – 2017-02-25
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    To put it in a [symmetric format](https://www.wolframalpha.com/input/?i=Plot%5B%7B%5B1%2Bx%5D%5E%5B1%2Bx%5D%2B%5B1-x%5D%5E%5B1-x%5D-2,3%5B1-%5B1+-+x%5E2%5D%5E%5B2%2F3%5D%5D%7D,%7Bx,-1,%2B1%7D%5D): $$ a=1+x,\,b=1-x \,\Rightarrow\, \color{blue}{(1+x)^{1+x}+(1-x)^{1-x}-2} \,\ge\, \color{red}{3\left[1-\left(1-x^2\right)^{2/3}\right]} \,\,\colon\,\, x\in[-1,\,+1] $$ – 2017-02-25
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    @Michael Rozenberg I have a proof of this inequality. It is neither short nor elegant (no significant insights), but only uses simple algebraic manipulations and properties of derivatives. Are you interested in such a proof? – 2017-03-28
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    @cafaxo Yes, of course! – 2017-03-28

1 Answers 1

2

We will show that if $x \in (-1,1)$, then $$(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5.$$

First we will prove that the inequality holds for all $x \in [-\frac{9}{10},\frac{9}{10}]$.

Lemma 1.1. If $x \in (-1,1)$, then $(1+x)^{1+x}\geq \frac{1}{12}x^5+\frac{1}{3}x^4+\frac{1}{2}x^3+x^2+x+1$.

Proof. For all $x \in (-1,1)$ we have \begin{align*} &\frac{1}{12}x^5+\frac{1}{3}x^4+\frac{1}{2}x^3+x^2+x+1 > 0 \\[7pt] &\impliedby \frac{1}{12}x^5+\frac{1}{3}x^4 > 0 \land \frac{1}{2}x^3 + x^2 > 0 \land x+1 > 0 \\[7pt] &\iff x^4 (x+4) > 0 \land x^2 (x+2) > 0 \land x+1 > 0, \end{align*} thus we can define $\gamma \colon (-1,1) \rightarrow \mathbb{R}$ by $$\gamma(x) = (1+x) \log (1+x) - \log\left(\frac{1}{12}x^5+\frac{1}{3}x^4+\frac{1}{2}x^3+x^2+x+1\right).$$ We show $\gamma \geq 0$. For all $x \in (-1,1)$ we have \begin{align*} &\gamma''(x) \geq 0 \\[7pt] &\iff \frac{x^4 \left(x^6+13 x^5+65 x^4+192 x^3+364 x^2+468 x+324\right)}{(x+1) \left(x^5+4 x^4+6 x^3+12 x^2+12 x+12\right)^2} \geq 0 \\[7pt] &\impliedby x^6+13 x^5+65 x^4+192 x^3+364 x^2+468 x+324 \geq 0 \\[7pt] &\impliedby x^6+13 x^5+65 x^4 \geq 0 \land 192 x^3+192 x^2 \geq 0 \land 172 x^2+468 x+324 \geq 0 \\[7pt] &\impliedby\operatorname{Discr} \left(x^2+13 x+65\right) = -91 \\[7pt] &\qquad\land x^2(x+1) \geq 0 \\[7pt] &\qquad\land \operatorname{Discr} \left(172 x^2+468 x+324\right) = -3888, \end{align*} therefore $\gamma'$ is increasing. Since $\gamma'(0) = 0$, we know that $\gamma'(x) \leq 0$ for all $x \in (-1,0]$ and $\gamma'(x) \geq 0$ for all $x \in [0,1).$ Thus $\gamma$ is decreasing on $(-1,0]$ and increasing on $[0,1)$. Since $\gamma(0) = 0$, we have $\gamma \geq 0$ and we are done. $$\tag*{$\Box$}$$

Lemma 1.2. If $x \in [-\frac{9}{10}, \frac{9}{10}]$ then $$8 x^8+72 x^6+108 x^4-432 x^2+243 \geq 0.$$

Proof. For all $x \in [-\frac{9}{10}, \frac{9}{10}]$ we have \begin{align*} &8 x^8 + 9 (8 x^6+12 x^4-48 x^2+27) \geq 0 \\ &\impliedby 8 x^6+12 x^4-48 x^2+27 \geq 0 \\ &\impliedby 8 x^6+12 x^4-48 x^2+27 + 4 \left(x^2-1\right)^3 \geq 0 \\ &\iff 12 (x^6-3 x^2+2)-1 \geq 0 \\ &\iff 12 (x^2-1)^2 \left(x^2+2\right) -1 \geq 0 \end{align*}

We define $\gamma \colon [0, \frac{9}{10}] \rightarrow \mathbb{R}$ by $$\gamma(x)=12 (x^2-1)^2 \left(x^2+2\right) -1.$$ For all $x \in [0, \frac{9}{10}]$ we have $$\gamma'(x) = 72 x(x^4-1) \leq 0,$$ thus $\gamma$ is decreasing. Since \begin{align*} &\gamma\left(\frac{9}{10}\right) \geq 0 \\ &\iff 12 \left(\frac{-19}{100}\right)^2 \cdot \frac{281}{100} -1 \geq 0 \\ &\iff \frac{12\cdot19^2 \cdot 281}{10^6} -1 \geq 0 \\ &\iff 12\cdot19^2 \cdot 281 \geq 10^6 \\ &\impliedby 19^2 \cdot 28 \geq 10^4 \\ &\iff 10108 \geq 10^4, \end{align*} we have $\gamma \geq 0$. By symmetry, we are done. $$\tag*{$\Box$}$$

Claim 1.3. If $x \in [-\frac{9}{10}, \frac{9}{10}]$ then $$(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5.$$

Proof. Let $x \in [-\frac{9}{10}, \frac{9}{10}]$. We have \begin{align*} &(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5 \\ \text{By Lemma 1.1:} \\ &\impliedby \left(\frac{1}{12}x^5+\frac{1}{3}x^4+\frac{1}{2}x^3+x^2+x+1\right) \\ &\qquad +\left(-\frac{1}{12}x^5+\frac{1}{3}x^4-\frac{1}{2}x^3+x^2-x+1\right) \\ &\qquad + 3 \left(1-x^2\right)^{2/3} \geq 5 \\[7pt] &\iff \frac{2}{3}x^4 + 2x^2 - 3 \geq -3 \left(1-x^2\right)^{2/3} \\[7pt] &\iff \left(\frac{2}{3}x^4 + 2x^2 - 3\right)^3 \geq -27 \left(1-x^2\right)^{2} \\[7pt] &\iff \frac{1}{27} x^4 \left(8 x^8+72 x^6+108 x^4-432 x^2+243\right) \geq 0. \\[7pt] &\impliedby \text{Lemma 1.2.} \end{align*} $$\tag*{$\Box$}$$

Now we will prove that the inequality holds for all $x \in [\frac{9}{10},1)$.

Lemma 2.1 If $x \in [\frac{1}{2},1]$ then $(1+x)^{1+x} \geq (4+4 \log 2)x-4 \log 2$.

Proof. Since for all $x \in [\frac{1}{2},1]$ we have \begin{align*} (4+4 \log 2)x-4 \log 2 &\geq (4+4 \log 2)\frac{1}{2}-4 \log 2 \\ &= 2-2 \log 2 \\ &> 2-2 \log \mathrm{e} = 0, \end{align*} we can define $\gamma \colon [\frac{1}{2},1] \rightarrow \mathbb{R}$ by $$\gamma(x) = (1+x) \log(1+x) - \log \left((4+4 \log 2)x-4 \log 2\right).$$ For all $x \in [\frac{1}{2},1]$ we have $$\gamma''(x) = \frac{1}{1+x}+\frac{(1+\log (2))^2}{(x+x \log (2)-\log (2))^2} \geq 0,$$ therefore $\gamma'$ is increasing. Since $\gamma'(1) = 0$, we know that $\gamma' \leq 0$. Thus $\gamma$ is decreasing. Since $\gamma(1) = 0$, we have $\gamma \geq 0$ and we are done. $$\tag*{$\Box$}$$

Claim 2.2 If $x \in [\frac{9}{10},1)$ then $$(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5.$$

Proof. For all $x \in [\frac{9}{10},1)$ we have $$(1-x)^{1-x} = \exp\left((1-x) \log(1-x)\right) \geq 1 + (1-x) \log(1-x),$$ therefore \begin{align*} &(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5 \\ &\impliedby (1+x)^{1+x}+1 + (1-x) \log(1-x)+3 \left(1-x^2\right)^{2/3} \geq 5 \\[10pt] &\text{By Lemma 2.1:} \\[10pt] &\impliedby (4+4 \log 2)x-4 \log 2+1 + (1-x) \log(1-x)+3 \left(1-x^2\right)^{2/3} \geq 5 \\ &\iff (4+4 \log 2)(x-1) + (1-x) \log(1-x)+3 \left(1-x^2\right)^{2/3} \geq 0 \\ &\iff (1-x)\left(-4-4 \log 2 + \log(1-x)+3(1-x)^{-1/3} (1+x)^{2/3}\right) \geq 0 \\ &\impliedby -4-4 \log 2 + \log(1-x)+3 (1-x)^{-1/3} (1+x)^{2/3} \geq 0. \end{align*} Let $\gamma\colon [\frac{9}{10},1) \rightarrow \mathbb{R}$ be the function given by $$\gamma(x) = \log(1-x)+3 (1-x)^{-1/3} (1+x)^{2/3} -4-4 \log 2.$$

For all $x \in [\frac{9}{10},1)$ we have \begin{align*} &\gamma'(x) \geq 0 \\ &\iff \frac{3-x-\left(1-x^2\right)^{1/3}}{(1-x)\left(1-x^2\right)^{1/3}} \geq 0 \\ &\impliedby 3-x-\left(1-x^2\right)^{1/3} \geq 0 \\ &\iff (3-x)^3 \geq 1-x^2 \\ &\iff -x^3+10 x^2-27 x+26 \geq 0 \\ &\impliedby -x^3+x^2 \geq 0 \land 9 x^2-27 x+18 \geq 0 \\ &\iff x^2(1-x) \geq 0 \land 9 (1-x)(2-x) \geq 0. \end{align*} Therefore $\gamma$ is increasing. Since \begin{align*} &\gamma\left(\frac{9}{10}\right) \geq 0 \\ &\iff \log\left(\frac{1}{10}\right)+3 \left(\frac{1}{10}\right)^{-1/3} \left(\frac{19}{10}\right)^{2/3} -4-4 \log 2 \geq 0 \\ &\iff -40+3\cdot 190^{2/3}-40 \log 2-10 \log 10 \geq 0 \\ &\impliedby -40+3\cdot 33-40 \log 2-10 \log 16 \geq 0 \\ &\iff \exp \frac{59}{80} \geq 2 \\ &\impliedby 1 + \frac{59}{80} + \frac{59^2}{2\cdot80^2} \geq 2 \\ &\iff 2\cdot59\cdot80 + 59^2 \geq 2\cdot80^2 \\ &\iff 12921 \geq 12800, \end{align*} we have $\gamma \geq 0$ and we are done. $$\tag*{$\Box$}$$

By symmetry we have $$(1+x)^{1+x}+(1-x)^{1-x}+3 \left(1-x^2\right)^{2/3} \geq 5$$ for all $x \in (-1,1)$.

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    @Michael: You should consider upvoting and accepting this impressive answer! – 2018-02-25