Bài 1:
a) Chứng minh [tex]\frac{a^{2}+b^{2}+c^{2}}{3}\geq (\frac{a+b+c}{3})^{2}[/tex]
b) [tex]\frac{x^{2}+y^{2}}{x-y}\geq 2\sqrt{2}[/tex]
c) a,b,c>0, chứng minh [tex]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}[/tex]
a, [tex]\frac{a^{2}+b^{2}+c^{2}}{3}\geq (\frac{a+b+c}{3})^{2}\\\\ <=> \frac{a^{2}+b^{2}+c^{2}}{3}\geq \frac{(a+b+c)^2}{9}\\\\ <=> a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}\\\\ <=> 3.(a^2+b^2+c^2)\geq (a+b+c)^2\\\\ <=> 3.(a^2+b^2+c^2)\geq a^2+b^2+c^2+2ab+2bc+2ca\\\\ <=> 2a^2+2b^2+2c^2-2ab-2bc-2ca\geq 0\\\\ <=> (a-b)^2+(b-c)^2+(c-a)^2\geq 0[/tex]
luôn đúng
dấu "=" xảy ra <=> a=b=c
b, sai đề nếu x<y => VT<0; mà VP>0 => vô lí
c, áp dụng Binhiacopxki có:
[tex]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\\\\ =\frac{a^2}{ab+ca}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\\\\ \geq \frac{(a+b+c)^2}{2.(ab+bc+ca)}[/tex]
lại có: [tex]a^2+b^2+c^2\geq ab+bc+ca\\\\ <=> a^2+b^2+c^2+2.(ab+bc+ca)\geq 3.(ab+bc+ca)\\\\ <=> (a+b+c)^2\geq 3.(ab+bc+ca)[/tex]
=> [tex]\frac{(a+b+c)^2}{2.(ab+bc+ca)}\geq \frac{(a+b+c)^2}{2.\frac{(a+b+c)^2}{3}}=\frac{3}{2}[/tex]
dấu "=" xảy ra <=> a=b=c