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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.

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    Do you have any thoughts? (Also, I have a pet peeve about people calling questions "hard" while giving as evidence only that they failed to do them.) – 2017-01-28
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    Ok !how can do? Change the subject? – 2017-01-28
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    The "edit" button underneath your post should let you do that. – 2017-01-28
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    Please edit the post to include more information: what is the source of the problem? What was its inspiration? Is it related to other problems? Is the inequality applicable in any area of math? This site is not intended for merely posting problems in the style of a contest, and posts the do nothing but state a problem are discouraged. The best posts are almost like blog posts, with enough information to make the problem broadly interesting. – 2017-01-28
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    @CarlMummert I try do it thanks for your help – 2017-01-28

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Let $a-1=x$ and $b-1=y$. Hence, by AM-GM $$\frac{a^4}{(b-1)^2}+\frac{b^4}{(a-1)^2}=\frac{(x+1)^4}{y^2}+\frac{(y+1)^4}{x^2}\geq\frac{16x^2}{y^2}+\frac{16y^2}{x^2}\geq32$$

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    Nice prove by placement on variables:) – 2017-01-28