# If a+b=1a+b=1 then a4b2+b4a2≤1a^{4b^2}+b^{4a^2}\leq1

Let $$aa$$ and $$bb$$ be positive numbers such that $$a+b=1a+b=1$$. Prove that:
$$a4b2+b4a2≤1a^{4b^2}+b^{4a^2}\leq1$$

I think this inequality is very interesting because the equality “occurs” for $$a=b=12a=b=\frac{1}{2}$$ and also for $$a→0a\rightarrow0$$ and $$b→1b\rightarrow1$$.

I tried to work with a function of one variable, but the derivative is not easy.

I also don’t get something solvable by Taylor series.

We define $$f(x,y)=x4y2+y4x2f(x,y)=x^{4y^2}+y^{4x^2}$$.

This is my plan to solve the problem:

• Since $$x+y=1x+y=1$$, we replace $$yy$$ by $$1−x1-x$$.
• We make a new function: $$g(x)=x4(1−x)2+(1−x)4x2g(x)=x^{4(1-x)^2}+(1-x)^{4x^2}$$
• Therefore, we must find the maximum on the range $$x∈[0,1]x \in [0,1]$$ of $$gg$$ so that we can see the maximum is less than or equal to $$11$$.

This will be troublesome:

$$g(x)=x4(1−x)2+(1−x)4x2g(x)=x^{4(1-x)^2}+(1-x)^{4x^2}$$

Set $$g1(x)=x4(1−x)2g_{1}(x) = x^{4(1-x)^2}$$ and $$g2(x)=(1−x)4x2g_{2}(x) = (1-x)^{4x^2}$$. Therefore, we can break it up like so:

$$g′(x)=g′1(x)+g′2(x)g'(x) = g_{1}'(x)+g_{2}'(x)$$
$$g′1(x)=g′1,g′2(x)=g′2g_{1}'(x)=g_{1}', g_{2}'(x)=g_{2}'$$
$$ln(g1)=ln(x4(1−x)2)\ln(g_{1})=\ln \left(x^{4(1-x)^2}\right)$$
$$ln(g1)=4(1−x)2⋅ln(x)\ln(g_{1})={4(1-x)^2} \cdot \ln \left(x\right)$$
$$g′1g1=4⋅((1−x)2)′⋅ln(x)+4(1−x)2x\frac{g_{1}'}{g_{1}}= 4 \cdot \left((1-x)^2 \right)' \cdot \ln(x)+\frac{4(1-x)^2}{x}$$
$$g′1g1=4⋅−2⋅(1−x)⋅ln(x)+4(1−x)2x\frac{g_{1}'}{g_{1}}= 4 \cdot -2 \cdot (1-x) \cdot \ln(x)+\frac{4(1-x)^2}{x}$$
$$g′1g1=8(x−1)ln(x)+4x2−8x+4x\frac{g_{1}'}{g_{1}}= 8(x-1)\ln(x)+\frac{4x^2-8x+4}{x}$$
$$g′1g1=8(x−1)ln(x)+4x−8+4x\frac{g_{1}'}{g_{1}}= 8(x-1)\ln(x)+4x-8+\frac{4}{x}$$
$$g′1=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)g_{1}'= x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right)$$

Alright. Deep breath. Let’s keep going.

$$ln(g2)=4x2ln(1−x)\ln(g_{2})=4x^2\ln(1-x)$$
$$g′2g2=8xln(x−1)+4x2x−1\frac{g_{2}'}{g_{2}}=8x\ln(x-1)+\frac{4x^2}{x-1}$$

$$g′2=(1−x)4x2(8xln(x−1)+4x2x−1)g_{2}'= (1-x)^{4x^2}\left(8x\ln(x-1)+\frac{4x^2}{x-1}\right)$$

$$g′1=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)g_{1}'= x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right)$$

$$g′(x)=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)+(1−x)4x2(8xln(1−x)+4x2x−1)g'(x)=x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right) + (1-x)^{4x^2}\left(8x\ln(1-x)+\frac{4x^2}{x-1}\right)$$

The maximum appears (according to the closed interval method), either at:

$$g(0)=1g(0)=1$$
$$g(1)=1g(1)=1$$

Or at the $$xx$$-value(s) of the solution of:

$$0=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)+(1−x)4x2(8xln(1−x)+4x2x−1)0=x^{4(1-x)^2} \cdot \left(8(x-1)\ln(x)+4x-8+\frac{4}{x}\right) + (1-x)^{4x^2}\left(8x\ln(1-x)+\frac{4x^2}{x-1}\right)$$

Therefore, if we set $$x1x_{1}$$, $$x2x_{2}$$, $$x3x_{3}$$ … to be the solutions to the equation above in the interval $$xn∈[0,1]x_{n} \in [0,1]$$, we have reduced the problem to proving that:

$$g(x1),g(x2),g(x3)...≤1g(x_{1}),g(x_{2}), g(x_{3})... \leq 1$$

Through some graphing of $$g(x)g(x)$$, we see that there exists $$x1x_{1}$$, $$x2x_{2}$$, and $$x3x_{3}$$, where $$x2x_{2}$$ is $$0.50.5$$ and the others are not easily calculatable or are irrational.

It can easily be seen that $$g′(0.5)=0g'(0.5) = 0$$ and that $$g(0.5)=1g(0.5)=1$$ (a maximum of the function). Since we now have proof that $$g(x2)≤1g(x_{2}) \leq 1$$ and we see that there does not exist an $$xnx_{n}$$ s.t. $$n>3n>3$$ and $$g′(xn)=0g'(x_{n})=0$$, we can reduce our previous problem to:

Prove that:

$$g(x1),g(x3)≤1g(x_{1}), g(x_{3}) \leq 1$$

Through Newton’s Method, we obtain approximations of $$x1x_{1}$$ and $$x3x_{3}$$ accurate to 10 decimal places. We state them below:

$$x1≈0.281731964017x_{1} \approx 0.281731964017$$
$$x3≈0.718268035983x_{3} \approx 0.718268035983$$

Note that:

$$g′(x1)≈g′(0.281731964017)=7.349676423⋅10−14g'(x_{1}) \approx g'(0.281731964017)=7.349676423 \cdot 10^{-14}$$
$$g(x1)≈g(0.281731964017)=0.973494223187g(x_{1}) \approx g(0.281731964017)=0.973494223187$$

We now have that $$g(x1)g(x_{1})$$ is a minimum of the function and that $$g(x1)≤1g(x_{1}) \leq 1$$

Finally:

$$g′(x3)≈g′(0.718268035983)=−7.349676423⋅10−14g'(x_{3}) \approx g'(0.718268035983)=-7.349676423 \cdot 10^{-14}$$
$$g(x3)≈g(0.718268035983)=0.973494223187g(x_{3}) \approx g(0.718268035983)=0.973494223187$$

We now have that $$g(x1)g(x_{1})$$ is also a minimum of the function and that $$g(x1)≤1g(x_{1}) \leq 1$$

We now have that:

$$g(x1),g(x2),g(x3)≤1g(x_{1}), g(x_{2}), g(x_{3}) \leq 1$$

Q.E.D

I took a very head-on brute-force approach to the problem, but I am happy with the rigorousness of the result and the final proof. We also now have the minimums of the function, which if anyone is curious, is $$≈0.973494223187\approx 0.973494223187$$