try for a gemetric proof!!!
7 Answers
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 :(
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 a1m+a2m+...+anm/n > (a1+a2+...+an/n)m if m belongs to R – [0, 1].
However if m belongs to (0, 1), then a1m+a2m+...+anm/n < (a1+a2+...+an/n)m
and if m {0, 1}, then a1m+a2m+...+anm/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