cho a,b,c,d > 0 c/m :[tex]\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\geqslant \frac{4}{3}[/tex]
Bổ đề : Cho [tex]x,y,z,t> 0[/tex]
[tex]\frac{a^2}{x}+\frac{b^{2}}{y}+\frac{c^{2}}{z}+\frac{d^{2}}{t}\geq \frac{(a+b+c+d)^2}{x+y+z+t}[/tex]
Áp dụng
[tex]\sum \frac{a}{b+c+d}=\sum \frac{a^2}{ab+ac+ad}\geq \frac{(a+b+c+d)^2}{\sum (ab+bc+cd+da)}[/tex]
Cần chứng minh
[tex]=\frac{(a+b+c+d)^2}{\sum (ab+ac+ad)}\\=\frac{a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd}{2ab+2ac+2ad+2bc+2bd+2cd}[/tex]
[tex]\geq \frac{4}{3}\\\Leftrightarrow 3(a^2+b^2+c^2+d^2)+3(2ab+2ac+2ad+2bc+2bd+2cd)\\\geq 4(2ab+2ac+2ad+2bc+2bd+2cd)\\\Leftrightarrow 3(a^2+b^2+c^2+d^2)\geq 2ab+2ac+2ad+2bc+2bd+2cd\\\Leftrightarrow (a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq 0[/tex]
(luôn đúng)
Vậy ta có đpcm
Dấu ''='' xảy ra khi a=b=c=d