# Nice inequality with exponents $a^{2b}+b^{2a}\leq a^{\Big(\frac{a(1-a)(\frac{1}{2}-a)}{4}\Big)^2}$

Hi it's a little refinement to play with a hard inequality of Vasile Cirtoaje :

Let $$a\geq b>0$$ such that $$a+b=1$$ then we have : $$a^{2b}+b^{2a}\leq a^{\Big(\frac{a(1-a)(\frac{1}{2}-a)}{4}\Big)^2}=f(a)$$

It implies directly the inequality of Vasile Cirtoaje .

I have tried the substitution $$a=\sinh^2(x)$$

$$\sinh(x)^{4\cosh^2(x)}+\cosh(x)^{4\sinh^2(x)}\leq \sinh(x)^{\Big(\frac{\cosh^2(x)\sinh^2(x)(\frac{1}{2}-\sinh^2(x))}{2}\Big)^2}$$

But I think it's nothing .

If we take one element of the sum and make the difference with the RHS and finally use derivatives it becomes awful . So I think it's a wrong way .

I have tried obviously Bernoulli's inequality as :

$$a^{2(1-a)}\leq 1+(a^2-1)((1-a)) \quad, (1-a)^{2(a)}\leq 1+((1-a)^2-1)(a)$$

But I don't know what to do next maybe there exists a reversed Bernoulli's inequality (?).Now I'm stuck because it's a hard nut (it could be my song).

## Little update

Maybe we can compare the upper bound got with Bernoulli's inequality with an inequality of the kind : $$1+\Big(\frac{a(1-a)(\frac{1}{2}-a)}{4}\Big )^{\alpha}\leq a^{\Big(\frac{a(1-a)(\frac{1}{2}-a)}{4}\Big)^2}=f(a)$$

We can determine easily $$\alpha$$ numericaly .

The inequality for $$0\le b\le 1/2$$, $$a+b=1$$, $$a^{2b}+b^{2a}\leq a^{\Big(\frac{ab(b-a)}{8}\Big)^2}\tag{1}$$ can be strengthened to $$a^{2b}+b^{2a}\le e^{-3a^2b^2(b-a)^2/4}\tag{2}$$ noting that $$e^{-3a^2b^2(b-a)^2/4}\le a^{\Big(\frac{ab(b-a)}{8}\Big)^2}$$ unless $$a$$ is almost $$0$$.

Only a sketch proof is given here, for what it's worth.

For brevity, let $$f(x):=x^{1-x}$$ and $$g(x):=x^2(1-x)^2(x-\frac{1}{2})^2$$. The claim is $$f(x)^2+f(1-x)^2\le e^{-3g(x)}$$

Trivially, $$f(x)\ge x$$, $$f(0)=0$$, $$f'(0)=1$$, $$f(1)=1$$, $$f'(1)=0$$. $$f'(x)=\frac{1-x-x\ln x}{x^x}$$.

Proposition 1. $$f'(x)=0\iff x=1$$

Proof: $$\frac{1}{x}-\ln x=1$$, equivalent to $$\frac{1}{x}+\ln\frac{1}{x}=1$$, so $$x=\frac{1}{W(e)}=1$$. ($$W$$ is Lambert's function.)

Proposition 2. $$f(x)^2+f(1-x)^2$$ has three local maxima, at $$x=0,\frac{1}{2},1$$.

Proof: The maxima/minima of $$f(x)^2+f(1-x)^2$$ occur when $$f(x)f'(x)=f(1-x)f'(1-x)$$.

At $$x=0$$, $$f(0)=0=f'(1)$$; at $$x=1$$, $$f(1-x)=0=f'(1)$$. Otherwise, divide by $$f(x)$$, $$f(1-x)$$.

At $$x=0$$, $$f(x)^2+f(1-x)^2=e^{2(1-x)\ln x}+e^{2x\ln(1-x)}=x^2+o(x)+1-2x^2+o(x)=1-x^2+o(x)$$ Hence $$x=0$$ is a local maximum. By symmetry, so is $$x=1$$.

The function $$\frac{f'(x)}{f(1-x)} = \frac{1-x-x \ln x}{(x(1-x))^x}$$ lies between $$1$$ and $$2$$, and has one local maximum and one local minimum. A sketch is as follows (blue curve). $$\frac{f'(x)}{f(1-x)}=\frac{f'(1-x)}{f(x)}$$ at three places; the intersections are simple. Since $$x=0$$ is a local maximum, it follows that the only other local maximum is at $$x=1/2$$. Then $$f(1/2)=1$$.

Corollary $$F(x):=-\ln(f(x)^2+f(1-x)^2)$$ also has three local minima at $$x=0,\frac{1}{2},1$$.

A Taylor expansion at each point gives $$F(0+h)=h^2+o(h)$$, $$F(1/2+h)=ch^2+o(h)$$ where $$c=4 - 4\ln2 - 2\ln^22\approx0.267$$, $$F(1-h)=h^2+o(h)$$.

Hence fitting a polynomial with double roots at $$x=0,\frac{1}{2},1$$, namely $$\alpha g(x)$$, a necessary condition for $$\alpha g(x)\le F(x)$$ is $$\alpha\le \min(4,16c)=4$$. A sketch of $$F(x)$$ shows that the worst cases are at these points; and that $$\alpha\le3$$ is sufficient for $$F(x)\ge3g(x)$$. No simple proof for this, just a splitting into ranges $$[0,1/8]$$, $$[1/8,3/8]$$, $$[3/8,1/2]$$, and use Taylor series on each. Proposition 3. For $$x>e^{-16\alpha}$$, $$e^{-\alpha g(x)}\le x^{(g(x)/4)^2}$$

Proof: Follows from $$-\alpha g(x)\le g(x)\ln x/16$$, equivalent to $$\ln x\ge-16\alpha$$, or $$x\ge e^{-16\alpha}$$.

• Thanks it's seems good to me but just for the proposition 3 you have a strict inequality wich becomes an equality .I think it's not a serious problem .(+1) – Erik Satie Jul 21 at 15:26
• Well in fact it's very nice I accept !! – Erik Satie Jul 21 at 15:27