phan gia huy
17/08/2017 at 21:57-
Perhaps Bunyakovsky's inequality :v , Sure right =)))
We have :
\(n.\left(a_1^2+a_2^2+.....+a_n^2\right)=\left(1+1+1+.......+1\right).\left(a_1^2+a_2^2.......+a_n^2\right)\)
........ n number 1..........
\(L.H.S\ge\left(1.a_1+1.a_2+.....+1.a_n\right)^2=\left(a_1+a_2+.....+a_n\right)^2\)
Done , ok !
Selected by MathYouLike -
a) \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(a^2+2ab+b^2\le2a^2+2b^2\)
\(a^2-2ab+b^2\ge0\)
\(\left(a-b\right)^2\ge0\)
b) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(a^2+b^2+c^2+2ab+2bc+2ca\le3a^2+3b^2+3c^2\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)