Suppose we have two squares $A$ and $B$, with surface area of
$$\big(\frac{a}{a+b}\big)^{2} , \big(\frac{b}{a+b}\big)^{2}$$
respectively. The sum of the surface areas of $A$ and $B$ is minimized when $a=b$. My question is why? What is the intuition behind this? Are there any explanations, or any references that could help me make sense of this intuitively?
By C-S $$\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{a+b}\right)^2=\frac{1}{2}\cdot\frac{(1^2+1^2)(a^2+b^2)}{a+b)^2}\geq\frac{1}{2}\cdot\frac{(a+b)^2}{(a+b)^2}=\frac{1}{2}.$$ The equality occurs for $(a,b)||(1,1)$, which happens for $a=b$.
Id est, $\frac{1}{2}$ is a minimal value.
Also, if you want to use your intuition then we have the following way.
For $a=b$ we get the value $\frac{1}{2}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that $$\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{a+b}\right)^2\geq\frac{1}{2}$$ or $$2(a^2+b^2)\geq(a+b)^2$$ or $$(a-b)^2\geq0,$$ which is obvious.
Done!