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 $a\rightarrow0^+$
On
WLOG, assume that $a \ge b$.
Case 1: $0 < b \le 1/8$
Let $f(x) := x^x$. Since $f(x)$ is convex on $x > 0$, we have $$a^a \ge f(2) + f'(2)(a - 2) = 4 + (4 + 4\ln 2)(a - 2) \ge 4 + 7(a - 2) = 7a - 10. $$
Also, we have $b^b = \mathrm{e}^{b\ln b} \ge 1 + b\ln b$.
It suffices to prove that $$7a - 10 + 1 + b\ln b + 3\sqrt[3]{a^2b^2} \ge 5$$ or $$-7b + b\ln b + 3\sqrt[3]{(2-b)^2b^2} \ge 0$$ or $$-7b^{1/3} + b^{1/3}\ln b + 3\sqrt[3]{(2 - b)^2} \ge 0.$$ Denote LHS by $g(b)$. We have $$g'(b) = - \frac{4}{3b^{2/3}} + \frac{\ln b}{3b^{2/3}} - \frac{2}{(2-b)^{1/3}} < 0.$$ Also, $g(1/8) > 0$. Thus, $g(b) > 0$ for all $0 < b \le 1/8$.
$\phantom{2}$
Case 2: $1/8 < b \le 1$
Fact 1: $u^u \ge \frac{u^2 - u + 2}{3 - u}$ for all $u$ in $(0, 2)$.
The proof is given at the end.
(Note: $\frac{u^2 - u + 2}{3 - u}$ is the Pade $(2,1)$ approximant of $u^u$ at $u = 1$.)
By Fact 1, it suffices to prove that $$\frac{a^2 - a + 2}{3 - a} + \frac{b^2 - b + 2}{3 - b} + 3\sqrt[3]{a^2b^2} \ge 5$$ or $$3\sqrt[3]{(2-b)^2b^2} \ge \frac{-11b^2 + 22b + 1}{(1+b)(3-b)}$$ or $$3^3{(2-b)^2b^2} \ge \left(\frac{-11b^2 + 22b + 1}{(1+b)(3-b)}\right)^3$$ which is true.
We are done.
Proof of Fact 1: Let $F(u) = u\ln u - \ln \frac{u^2 - u + 2}{3 - u}$. We have \begin{align*} F'(u) &= \ln u - \frac{(u - 1)(u^2 - 4u + 7)}{(3 - u)(u^2 - u + 2)},\\ F''(u) &= \frac{(u^4 - 5u^3 + 3u^2 - 19u + 36)(u - 1)^2}{u(3 - u)^2(u^2 - u + 2)^2}. \end{align*} Let $h(u) = u^4 - 5u^3 + 3u^2 - 19u + 36$. We have $h'(u) = u^2(4u - 15) + (6u - 19) < 0$. Thus, $h(u)$ is strictly decreasing on $(0, 2)$. Also, $h(1) > 0$ and $h(2) < 0$. Thus, there exists $u_0\in (1, 2)$ such that $h(u_0) = 0$, $h(u) > 0$ on $(0, u_0)$, and $h(u) < 0$ on $(u_0, 2)$. Thus, $F''(u)\ge 0$ on $(0, u_0]$ and $F''(u) < 0$ on $(u_0, 2)$. Thus, $F(u)$ is convex on $(0, u_0]$ and concave on $(u_0, 2)$. Since $F(1) = 0$ and $F'(1) = 0$, we have $F(u)\ge 0$ on $(0, u_0]$. Also, since $F(u_0)\ge 0$ and $F(2) = 0$, we have $F(u)\ge 0$ on $(u_0, 2)$. Thus, $F(u)\ge 0$ on $(0, 2)$. We are done.
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 $x \in (-1,1)$.