Let a and b be positive numbers such that a+b=1. Prove that:

a4b2+b4a2≤1I think this inequality is very interesting because the equality “occurs” for a=b=12 and also for a→0 and b→1.

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.

**Answer**

We define f(x,y)=x4y2+y4x2.

This is my plan to solve the problem:

- Since x+y=1, we replace y by 1−x.
- We make a new function: g(x)=x4(1−x)2+(1−x)4x2
- Therefore, we must find the maximum on the range x∈[0,1] of g so that we can see the maximum is less than or equal to 1.

This will be troublesome:

g(x)=x4(1−x)2+(1−x)4x2

Set g1(x)=x4(1−x)2 and g2(x)=(1−x)4x2. Therefore, we can break it up like so:

g′(x)=g′1(x)+g′2(x)

g′1(x)=g′1,g′2(x)=g′2

ln(g1)=ln(x4(1−x)2)

ln(g1)=4(1−x)2⋅ln(x)

g′1g1=4⋅((1−x)2)′⋅ln(x)+4(1−x)2x

g′1g1=4⋅−2⋅(1−x)⋅ln(x)+4(1−x)2x

g′1g1=8(x−1)ln(x)+4x2−8x+4x

g′1g1=8(x−1)ln(x)+4x−8+4x

g′1=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)

Alright. Deep breath. Let’s keep going.

ln(g2)=4x2ln(1−x)

g′2g2=8xln(x−1)+4x2x−1

g′2=(1−x)4x2(8xln(x−1)+4x2x−1)

g′1=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)

g′(x)=x4(1−x)2⋅(8(x−1)ln(x)+4x−8+4x)+(1−x)4x2(8xln(1−x)+4x2x−1)

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

g(0)=1

g(1)=1

Or at the x-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)

Therefore, if we set x1, x2, x3 … to be the solutions to the equation above in the interval xn∈[0,1], we have reduced the problem to proving that:

g(x1),g(x2),g(x3)...≤1

Through some graphing of g(x), we see that there exists x1, x2, and x3, where x2 is 0.5 and the others are not easily calculatable or are irrational.

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

Prove that:

g(x1),g(x3)≤1

Through Newton’s Method, we obtain approximations of x1 and x3 accurate to 10 decimal places. We state them below:

x1≈0.281731964017

x3≈0.718268035983

Note that:

g′(x1)≈g′(0.281731964017)=7.349676423⋅10−14

g(x1)≈g(0.281731964017)=0.973494223187

We now have that g(x1) is a minimum of the function and that g(x1)≤1

Finally:

g′(x3)≈g′(0.718268035983)=−7.349676423⋅10−14

g(x3)≈g(0.718268035983)=0.973494223187

We now have that g(x1) is also a minimum of the function and that g(x1)≤1

We now have that:

g(x1),g(x2),g(x3)≤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

**Attribution***Source : Link , Question Author : Michael Rozenberg , Answer Author : Maximilian Janisch*