Exercise 1.10 [Section 1, BOMC-2011]

[MATHPRITOM]

let,$x \gg y$ & then,using rearrangement inequality gives a nice solution,i think.

[*Mahi*]

Hmm.
Let $x>y$. Then $\frac 1 {\sqrt x} \leq \frac 1 {\sqrt {y}} $
So, by rearrangement inequality, $\frac x {\sqrt y}+ \frac y {\sqrt x} \geq \frac x {\sqrt x}+ \frac y {\sqrt y} $

[rakeen]

If $x,y > 0$,then $\sqrt {\frac{x^2}{y}}+ \sqrt{\frac{y^2}{x}} \geq \sqrt x + \sqrt y$

we can simplify the left side as $x \frac{1}{\sqrt {y}} + y \frac{1}{\sqrt {x}}$
now we can conject that, that term is greater than $\sqrt {x} + \sqrt {y}$ because x,y are non negetive and $\frac {1}{\sqrt {a}}$ is very small; moreover we have $\sqrt {x^2}$ . so the inequality is visibly true! but how can I prove it mathematially? can I use contradiction technique here?
-rakeen

[sourav das]

The term is obviously not visibly true.
Because as\[ x \to \infty ;\sqrt{x} \to \infty ;\frac{1}{\sqrt{x}} \to 0\]
So visibly we are confused. And there are many thing which are visibly true but actually not true. A great example Theory of Relativity . You may thought earth is still but that doesn't mean it really is!! So we have to prove something to be true other wise we are actually blind. This is why science is interesting.