chalte chalte prove inequality

try for a gemetric proof!!!

Are you sure about what you have written?

Grandmaster "probably" made a typo.
what he meant may be: \frac{{x_1}^2+{x_2}^2}{2}\geq \frac{(x_1+x_2)^2}{4}

this is the well known cauchy schwartz inequality...

@jake yes.....this is waht i meant to type......should have used the latex :(

snybody willing to prove it!!!!!

A.M of mth powers> mth power of A.M.

\frac{\sum{a_i^n}}{m}\ge \left(\frac{\sum{a_i}}{m} \right)^n, for m>1.

Let a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number. Then a1^m+a2^m+...+an^m/n > (a1+a2+...+an/n)^m if m belongs to R – [0, 1].

However if m belongs to (0, 1), then a1^m+a2^m+...+an^m/n < (a1+a2+...+an/n)^m

and if m {0, 1}, then a1^m+a2^m+...+an^m/n = (a1+a2+...+an/n)^m.

the above is
(x1-x2)^2\geq0.......(since both are real)

x1^2+x2^2\geq x1.x2+x1.x2

x1^2+x2^2+x1^2+x2^2\geq x1^2+x2^2+x1.x2+x1.x2

\frac{x1^2+x2^2+x1^2+x2^2}{4}\geq \frac{x1^2+x2^2+x1.x2+x1.x2}{4}

\frac{x1^2+x2^2}{2}\geq (\frac{x1+x2}{2})^2