If $a,b>1$ prove: $$\frac{a^4}{(b-1)^2}+\frac{b^4}{(a-1)^2}\ge32$$ I try to prove it by AM-GM but still working on it but there's a solution with placement variables.
I think solving this problem is worthy as well as this problem: $ x,y $ are positive real and$ x+y=1 $prove: $$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}\geq\sqrt{2}$$
This problem is easy with Jensen inequality but there is beautiful solution with AM-GM. The second problem is from India.
Let $a-1=x$ and $b-1=y$. Hence, by AM-GM $$\frac{a^4}{(b-1)^2}+\frac{b^4}{(a-1)^2}=\frac{(x+1)^4}{y^2}+\frac{(y+1)^4}{x^2}\geq\frac{16x^2}{y^2}+\frac{16y^2}{x^2}\geq32$$