# Prove that $a^{4b}+b^{4a}\geq \frac{1}{2}$

Inspired by a problem of Vasile Cirtoaje I propose this :

Let $$a,b>0$$ such that $$a+b=1$$ then we have : $$a^{4b}+b^{4a}\geq \frac{1}{2}$$

I compute the derivative of $$f(x)=x^{4(1-x)}+(1-x)^{4x}$$ on $$]0,1]$$ we get : $$f'(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x)) + (1 - x)^{(4 x)} (4 \log(1 - x) - \frac{4 x}{1 - x})$$

If we denote by $$g(x)$$ the function :

$$g(x)=x^{4 (1 - x)} (\frac{4 (1 - x)}{x} - 4 \log(x))$$

We can rewrite the derivative as :

$$f'(x)=g(x)-g(1-x)$$

So it's remains to show that $$g(x)\geq g(1-x)$$ or $$g(x)\leq g(1-x)$$

So it remains to show that $$g(x)$$ is increasing or decreasing .

After that I'm stuck...

Any helps are very appreciated !

Thanks a lot for your time .

Alternative Proof:

We need to prove that, for all $$a$$ in $$(0, 1/2]$$, $$a^{4 - 4a} + (1 - a)^{4a} \ge 1/2.$$

We split into two cases:

Case I $$\,\, a\in (0, 1/4)$$:

We have $$a^{4 - 4a} + (1 - a)^{4a} \ge (1 - a)^{4a} \ge 1 - a > \frac12.$$

Case II $$\,\,a \in [1/4, 1/2]$$:

First, using Bernoulli inequality, we have \begin{align*} a^{4 - 4a} &= 2^{-1}2^{-(2 - 4a)}(1 - (1 - 2a))^{3 - 4a} a\\ &\ge 2^{-1}2^{-(2 - 4a)}[1 - (1 - 2a)(3 - 4a)] a\\ &\ge 2^{-1}[1 - (2 - 4a)\ln 2]\,[1 - (1 - 2a)(3 - 4a)]a \end{align*} where we have used $$2^{-(2 - 4a)} = \mathrm{e}^{-(2 - 4a)\ln 2} \ge 1 - (2 - 4a)\ln 2$$.

Second, using Bernoulli inequality, we have \begin{align*} (1 - a)^{4a} &= 2^{-2}2^{2 - 4a}(1 + (1 - 2a))^{4a}\\ &\ge 2^{-2}2^{2 - 4a}[1 + (1 - 2a)\cdot 4a]\\ &\ge 2^{-2}[1 + (2 - 4a)\ln 2]\,[1 + (1 - 2a)\cdot 4a] \end{align*} where we have used $$2^{2 - 4a} = \mathrm{e}^{(2 - 4a)\ln 2} \ge 1 + (2 - 4a)\ln 2$$.

It suffices to prove that \begin{align*} &2^{-1}[1 - (2 - 4a)\ln 2]\,[1 - (1 - 2a)(3 - 4a)]a\\ &\quad + 2^{-2}[1 + (2 - 4a)\ln 2]\,[1 + (1 - 2a)\cdot 4a] \ge \frac12 \end{align*} or $$(1 - 2a)^2[-16(\ln 2)a^2 + (20\ln 2 - 4)a + 2\ln 2 - 1] \ge 0$$ which is true.

We are done.

Alternative proof:

I give a proof following @Malper's nice idea.

Fact 1: Let $$x\in (0, 1/2]$$ and $$t \in (0, 1/2]$$. Then $$x^{4 - 4t} + (1 - x)^{4t} \ge \frac12$$. The proof is given at the end.

By Fact 1, we have $$x^{4 - 4x} + (1 - x)^{4x} \ge \frac12$$ for all $$x$$ in $$(0, 1/2]$$.

We are done.

Proof of Fact 1: Let $$f(x, t) = x^{4 - 4t} + (1 - x)^{4t}$$. We have \begin{align} \frac{\partial f}{\partial t} &= 4(1 - x)^{4t}\ln (1 - x) - 4x^{4 - 4t}\ln x\\ &= 4x^{-4t}\Big([x(1 - x)]^{4t}\ln(1 - x) - x^4\ln x\Big)\\ &\le 4x^{-4t}\Big([x(1 - x)]^{4\cdot \frac12}\ln(1 - x) - x^4\ln x\Big)\\ &= 4x^{2 - 4t}\Big((1 - x)^2\ln(1 - x) - x^2\ln x\Big)\\ &\le 0 \end{align} where we have used $$(1 - x)^2\ln(1 - x) - x^2\ln x \le 0$$.
Note: Let $$g(x) = (1 - x)^2\ln(1 - x) - x^2\ln x$$. We have $$g''(x) = 2\ln(1 - x) - 2\ln x \ge 0$$ on $$(0, 1/2]$$. Thus, $$g(x)$$ is convex on $$(0, 1/2]$$. Also, $$g(0^{+}) = 0$$ and $$g(1/2) = 0$$. Thus, $$g(x) \le 0$$ on $$(0, 1/2]$$.

Thus, we have $$f(x, t) \ge f(x, 1/2) = x^2 + (1 - x)^2 \ge 1/2$$.

We are done.