\(\text{Σ}\frac{a}{b+2c+3d}=\text{Σ}\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{6\left(ab+bc+cd+ad\right)}\)
\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}=\frac{a^2+c^2+b^2+d^2+2ab+2cd+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}\)
\(\ge\frac{4\left(ab+bc+cd+ad\right)}{6\left(ab+bc+cd+ad\right)}=\frac{2}{3}\)
Dấu = xảy ra khi a=b=c=d
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4.\left(ab+ad+bc+bd+ca+cd\right)}\)\(\ge\frac{\left(a+b+c+d\right)^2}{\frac{3}{2}.\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d\)
\(VT=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
Áp dụng BĐT Svac-xơ cho 3 số dương ta được :
\(VT\ge\frac{\left(a+b+c+d\right)^2}{4ab+4ac+4ad+4bc+4bd+4cd}\)
Áp dụng BĐT phụ \(x^2+y^2\ge2xy\) ta được :
\(a^2+b^2\ge2ab;a^2+c^2\ge2ac;a^2+d^2\ge2ad\)
\(b^2+c^2\ge2bc;b^2+d^2\ge2bd;c^2+d^2\ge2cd\)
\(\Rightarrow3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)
Ta lại có : \(\left(a+b+c+d\right)^4=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\)
\(\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{3}\)
\(\Rightarrow VT\ge\frac{\left(a+b+c+d\right)^4}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{12\left(ab+ac+ad+bc+bd+cd\right)}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=d\)
