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It's a refinement of If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$

Let $a,b>0$ such that $a+b=1$ then we have : $$a^{4b^2}+b^{4a^2}\leq (a^{2b}+b^{2a})^{ab}\leq 1$$

The RHS is well-know . I try to use power series ($x=0$) on the following function ($x\in[0,1]$):

$$f(x)=(1-x)^{4x^2}+x^{4(1-x)^2}-(x^{2(1-x)}+(1-x)^{2x})^{x(1-x)}$$

We get : $$\begin{align} &-3 x^3 + x^4 (2 \log(x) - 1) + x^5 (-2 \log^2(x) - 10 \log(x) - 19/6)\\ &+ x^6 ((4 \log^3(x))/3 + 34 \log^2(x) + 6 \log(x) + 41/6)\\ &+ x^7 (-2/3 \log^4(x) - (260 \log^3(x))/3 - 32 \log^2(x) - 2 \log(x) + 71/10)\\ &+ 1/15 x^8 (4 \log^5(x) + 2570 \log^4(x) + 1880 \log^3(x) + 90 \log^2(x) - 10 \log(x) + 119)\\ &+ O(x^9) \end{align}$$ But I think we are in the wrong way maybe we can be inspired by this https://link.springer.com/article/10.1186/1029-242X-2013-468

So If you have nice idea it would be cool

Thanks a lot for sharing your time and knowledge .

Edit :

My help

Following the work of River Li we have :

Let $a\geq b>0$ such that $a+b=1$ and $b\in [0.3,0.5]$ then we have : $$a^{4b^2}+b^{4a^2}=(1-2ab-a^2)^{2a^2}+(1-2ab-b^2)^{2b^2}\leq 2\Big(1-2ab-\frac{a^2+b^2}{2}\Big)^{a^2+b^2}\quad (E)$$ And $$ 2\Big(1-2ab-\frac{a^2+b^2}{2}\Big)^{a^2+b^2}= 2\Big(\frac{a^2+b^2}{2}\Big)^{a^2+b^2}\leq (a^{2b}+b^{2a})^{a^2+b^2}$$ And $$(a^{2b}+b^{2a})^{a^2+b^2}\leq (a^{2b}+b^{2a})^{ab}$$

We have also :

Let $a\geq b>0$ such that $a+b=1$ and $b\in [0.3,0.5]$ then we have : $$a^{4b^2}+b^{4a^2}\leq a^{2b^2+b}+b^{2a^2+a}\leq a^{2b}+b^{2a}\leq (a^{2b}+b^{2a})^{ab}$$

A suggestion

For the inequality $(E)$ one can use Jensen's inequality with the concavity of the function :

$$f(x)=(\alpha-x)^{2x}$$

Where $\alpha=\operatorname{constant}<1$ and $\sqrt{x}\in[0.3,0.5]$

An other way

One can show that the function :

$$f(x)=x^{4(1-x)^2}+(1-x)^{4x^2}-(x^{2(1-x)}+(1-x)^{2x})$$

Is increasing for $x\in [0.3,0.5]$

It's miraculous because : $$f'(0.3)=0.0052865\cdots$$

That's all for me .

Thanks again.

Second edit :

Since the function (where $0\leq x \leq 1$ and $a+b=1$ and $a,b>0$):

$$f(x)=a^{4b^2(1-x)+2bx}+b^{4a^2(1-x)+2ax}$$

Is convex we have by Jensen's inequality :

$$f(0)+f(1)\geq 2f(0.5)$$

Or :

$$a^{4b^2}+b^{4a^2}+a^{2b}+b^{2a}\geq 2(a^{2b^2+b}+b^{2a^2+a})$$

We just need to show :

Let $a\geq b>0$ such that $a+b=1$ and $b\in [0.3,0.5]$ then we have : $$a^{4b^2}+b^{4a^2}\leq a^{2b^2+b}+b^{2a^2+a}$$

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Update

It remains to prove the case when $b \in [\frac{3}{10}, \frac{1}{2}]$.

From Proposition 5.2 in [1], we have $a^{2b} + b^{2a} \le 1$. Since $a^{2b} + b^{2a} \le (a^{2b} + b^{2a})^{ab}$, it remains to prove the following results (see The.old.crap's work):

Claim 1: Let $a = 1-b$ and $b\in [\frac{3}{10}, \frac{1}{2}]$. Then $a^{4b^2} + b^{4a^2} \le a^{2b} + b^{2a}$.

[1] Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions", The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137. https://eudml.org/doc/223938

Partial answer

Problem: Let $a, b > 0$ with $a+b=1$. Prove that $$a^{4b^2} + b^{4a^2} \le (a^{2b} + b^{2a})^{ab}.$$

WLOG, assume that $a\ge b$. Then we have $a = 1- b$ and $b\in (0, \frac{1}{2}]$.

First, let us prove the case when $b\in (0, \frac{3}{10}]$.

We have the following auxiliary results (Facts 1 through 6). The proof of Fact 5 is given later. For the proof of Fact 1, see How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$

Fact 1: $u^v \ge \frac{u}{u+v-uv}$ for $u>0, \ v\in [0, 1]$.

Fact 2: By using Fact 1, $(1-b)^{2b} \ge \frac{1-b}{2b^2 - b + 1}$ for $b\in (0, \frac{1}{2}]$.

Fact 3: By using Fact 1, $b^{2(1-b)} = b \cdot b^{1-2b} \ge \frac{b^2}{2b^2 - 2b + 1}$ for $b\in (0, \frac{1}{2}]$.

Fact 4: By using Bernoulli's inequality, we have $(1-b)^{4b^2} \le 1 - 4b^3$ for $b\in (0, \frac{1}{2}]$.

Fact 5: $b^{-8b + 4b^2} \le 12 - \frac{2}{3}b$ for $b\in (0, \frac{1}{2}]$.

Fact 6: By using Fact 5, $b^{4(1-b)^2} = b^4 \cdot b^{-8b + 4b^2} \le b^4(12 - \frac{2}{3}b)$ for $b\in (0, \frac{1}{2}]$.

From Facts 1, 2 and 3, we have \begin{align} (a^{2b} + b^{2a})^{ab} &= ((1-b)^{2b} + b^{2(1-b)})^{b(1-b)}\\ &\ge \left(\frac{1-b}{2b^2 - b + 1} + \frac{b^2}{2b^2 - 2b + 1}\right)^{b(1-b)}\\ &= w^{b(1-b)}\\ &\ge \frac{w}{w + b(1-b) - wb(1-b)}\\ &= \frac{2b^4-3b^3+5b^2-3b+1}{-2b^6+5b^5-2b^4-2b^3+5b^2-3b+1} \end{align} where $w = \frac{1-b}{2b^2 - b + 1} + \frac{b^2}{2b^2 - 2b + 1}$ (Clearly $w>0$ and $b(1-b)\in [0,1)$). With this in mind, from Facts 4, 5 and 6, it suffices to prove that for $b\in (0, \frac{3}{10}]$, $$1-4b^3 + b^4(12 - \tfrac{2}{3}b) \le \frac{2b^4-3b^3+5b^2-3b+1}{-2b^6+5b^5-2b^4-2b^3+5b^2-3b+1}$$ or $$\frac{b^3 (-4 b^8+82 b^7-208 b^6+128 b^5+58 b^4-204 b^3+155 b^2-60 b+9)}{-6 b^6+15 b^5-6 b^4-6 b^3+15 b^2-9 b+3} \ge 0.$$ It is not hard.

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Proof of Fact 5: It suffices to prove that for $b\in (0, \frac{1}{2}]$, $$\ln (12 - \tfrac{2}{3}b) \ge (-8b+4b^2)\ln b.$$ It is easy to prove that for $(0, \frac{1}{2}]$, $$\ln (12 - \tfrac{2}{3}b) \ge \frac{25539}{10325} - \frac{10}{177}b.$$ Thus, it suffices to prove that for $(0, \frac{1}{2}]$, $$f(b) = \frac{\frac{25539}{10325} - \frac{10}{177}b}{8b-4b^2} + \ln b \ge 0.$$ We have $$f'(b) = \frac{(10b-3)(6195b^2-23009b+25539)}{61950b^2(2-b)^2}.$$ Thus, $f(b)$ is strictly decreasing on $(0, \frac{3}{10})$, and strictly increasing on $(\frac{3}{10}, \frac{1}{2}]$. Also, we have $f(\frac{3}{10}) > 0$. The desired result follows.

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  • $\begingroup$ For your fact 1 I think (maybe) we can use the reverse Young's inequality .Furthermore I have a question : can we avoid the derivative in this problem ? $\endgroup$
    – Erik Satie
    Dec 22, 2019 at 12:46
  • $\begingroup$ One can make use of Bernoulli's inequality to prove Fact 1. $\endgroup$
    – River Li
    Dec 22, 2019 at 13:52

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