Lopsided ugly inequality

$\frac{(m-n)^2}{m(\frac{m+n}{2}-\sqrt{mn})}= \frac{2(m-n)^2}{m(m+n-2\sqrt{mn})}= \frac{2(m-n)^2}{m(\sqrt{m}-\sqrt{n})^2}= \frac{2(\sqrt{m}+\sqrt{n})^2}{m}$

$=2(\sqrt{\frac{n}{m}}+1)^2 \leq 2(\sqrt{\frac{n}{n+1}}+1)^2$ (since $m>n$ and they are positive integers)

Now, note that $\frac{n}{n+1}<\frac{n+1}{n+2}$ for positive integers $n$ and $\lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = 1$ . Hence

$\frac{(m-n)^2}{m(\frac{m+n}{2}-\sqrt{mn})}<2(1+1)^2=8$ , which is the smallest value of $x$ .

The value of $a+b$ is therefore $8+1=9$ .

$\frac{(m-n)^2}{m(\frac{m+n}{2}-\sqrt{mn})}$ $= \frac{(m-n)^2}{m(\frac{m+n-2\sqrt{mn}}{2})}$ $= \frac{(m-n)^2}{m(\frac{(\sqrt{m}-\sqrt{n}^2)}{2})}$ $= \frac{2(m-n)^2}{m(\sqrt{m}-\sqrt{n})^2}$ $= (\frac{2}{m})(\frac{(m-n)^2}{(\sqrt{m}-\sqrt{n})^2})$ $= (\frac{2}{m})(\sqrt{m}+\sqrt{n})^2$ $=(\frac{2}{m})(m+n+2\sqrt{mn})$ $=2+2\frac{n}{m}+4\sqrt{\frac{n}{m}}$ .

We let $\sqrt{\frac{n}{m}}=k$ , so the expression becomes $2+2k^2+4k=2k^2+4k+2$ . Since $k$ could be any positive real number lesser than $1$ by the conditions of the problem, $2k^2+4k+2<2(1)^2+4(1)+2=8$ for any $k$ as for $0<k<1$ , $k^2, k <1$ . Since for any $m, n$ we have a corresponding $k$ , the smallest $x$ such that $\frac{(m-n)^2}{m(\frac{m+n}{2}-\sqrt{mn})} <x$ is true is for all $(m,n)$ is $8=\frac{8}{1}$ .

Thus, the answer is $8+1=\boxed{9}$ .

Observe $\frac{(m-n)^2}{m( \frac{m+n}{2} - \sqrt{mn} )} = \frac{2(\sqrt{m} + \sqrt{n})^2}{m} \leq \frac{2(2\sqrt{m})^2}{m} = 8$ .

$\square$

We can write $\frac{m+n}{2}-\sqrt{mn}$ as $\frac{1}{2} \times (\sqrt{m}-\sqrt{n})^2$ . This turns the given expression to $\frac {(m-n)^2} {\frac{m}{2} (\sqrt{m}-\sqrt{n})^2 }$ which we can then further simplify to $\frac {2(m-n)^2} {m (\sqrt{m}-\sqrt{n})^2 }$ Now we can write $(m-n)=(\sqrt{m}-\sqrt{n})(\sqrt{m}+\sqrt{n})$ and subsitute into the resulting expression to get $\frac {2(\sqrt{m}-\sqrt{n})^2(\sqrt{m}+\sqrt{n})^2} {m (\sqrt{m}-\sqrt{n})^2 }$ . This simplifies to $\frac {2(\sqrt{m}+\sqrt{n})^2} {m }$ Writing $m=\sqrt{m}\times\sqrt{m}$ we can further rearrange the expression to give us $2(1+\sqrt{ \frac{n}{m} } )^2$ . Since we know that $\frac{n}{m}<1$ , we know that the entire expression is less than $2\times(1+1)^2=x=\fbox{8}$

(m−n)2m(m+n2−mn√)=2(m−n)2m(m+n−2mn√)=2(m−n)2m(m√−n√)2=2(m√+n√)2m =2(nm−−√+1)2≤2(nn+1−−−√+1)2 (since m>n and they are positive integers)

Now, note that nn+1<n+1n+2 for positive integers n and limn→∞nn+1−−−√=1. Hence

(m−n)2m(m+n2−mn√)<2(1+1)2=8, which is the smallest value of x.

The value of a+b is therefore 8+1=9.

[Staff Edit- This user has been removed for plagiarism]