For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$

Question

For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$

I couldn't solve this question and hence looked at the solution. The solution substitutes $x=\sqrt{a}$, $y=\sqrt{b}$, $z=\sqrt{c}$, hence $x^3+y^3+z^3=3$. We also have that $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le24$

which is also written as $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le \frac{8}{3}(x^3+y^3+z^3)^2$ which holds true, after expanding etc.

My question is, is there a simpler and more intuitive approach? And if not how was I supposed to think of the substitution used above?

Answer 1

Here I will show the thoughts that could help you to do all the steps above.

1. The first substitution is unnecessary, but it simplifies the algebra, since you have $(\sqrt{a})^3+(\sqrt{b})^3+(\sqrt{c})^3=3$
2. Now, for the inequality we want to prove we have $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3$ on the left side. The general trick is to make inequality Homogeneous. It means that if you imagine $x,y,z$ to be lengths in meters, you will get $meters^6$ on the left side, which means that we want to get $meters^6$ on the right side as well. Since $(x^3+y^3+z^3)$ is $meters^3$, we want to square it to get $meters^6$.
3. The last step is just to find the coefficient before $(x^3+y^3+z^3)^2$. Here it helps to check the equality point $(x^3+y^3+z^3)^2=9$, but we want to have 24. Therefore we put $\frac{8}{3}$ as a coefficient and get $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3 \leq \frac{8}{3}(x^3+y^3+z^3)^2$.