Lopsided ugly inequality

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 $2\times(1+1)^2=x=\fbox{8}$

Given, $\frac{(m-n)^2}{m(\frac{(m+n}{2}-\sqrt{mn})}<x$

$or,\frac{(\sqrt{m}+\sqrt{n})^2}{\frac{m}{2}}<x$

[Dividing nominator and denominator on L.H.S by $(\sqrt{m}-\sqrt{n})^2$

since, $m>n$ ,

hence, $(\sqrt{m}-\sqrt{n})\neq0$ ]

$or,(1+\sqrt{\frac{n}{m}})^2<\frac{x}{2}............(i)$

[Dividing nominator and denominator on L.H.S by m

since,m>n and n is positive integer then $m\neq0$ ]

but, $\frac{n}{m}<1 (since,m>n)$

hence, $\sqrt{\frac{n}{m}}<1$

then, $(1+\sqrt{\frac{n}{m}})<2$

$or,(1+\sqrt{\frac{n}{m}})^2<4............(ii)$

compairing (i) and (ii) we can say that $\frac{x}{2}=4$

hence,x=8 then a=8,b=1

(a+b)=9 (Ans)

Make the substitution $m = n + k$ , where $k$ is a positive integer. By multiplying both the numerator and denominators by the sum of the two means and exploiting the difference of squares identity in the denominator, you get the following rational expression in n and k: $\frac{2 \left( 2n + k + 2 \sqrt{n^2 + kn} \right)}{n + k}.$

The value of this expression decreases with $k$ and increases with $n$ . Hence, put $k = 1$ , and evaluate the limit as $n \to \infty$ . The limit is $8 = \frac{8}{1}$ , so the answer is $8 + 1 = 9.$

$\frac{(m-n)^2}{m(\frac{m+n}{2}-\sqrt{mn})}= \frac{2(m-n)^2}{m(m+n-2\sqrt{mn})}= \frac{2(m-n)^2}{m(\sqrt{m}-\sqrt{n})^2}= \frac{2(\sqrt{m}+\sqrt{n})^2}{m}$ $=2(\sqrt{\frac{n}{m}}+1)^2 \leq 2(\sqrt{\frac{n}{n+1}}+1)^2$ (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 $8+1=9$ .

Simplifying we get, LHS = 2[1 + 2sqr(n/m) + (n/m)] So, n/m must be a perfect square; Here, given m>n and for large consecutive perfect square n/m is nearly equal to 1. So, least value of x will be 2(1+2+1) = 8. So, a = 8 and b = 1.