\(\text{Σ}\frac{a}{b+2c+3d}=\text{Σ}\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{6\left(ab+bc+cd+ad\right)}\)

\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}=\frac{a^2+c^2+b^2+d^2+2ab+2cd+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}\)

\(\ge\frac{4\left(ab+bc+cd+ad\right)}{6\left(ab+bc+cd+ad\right)}=\frac{2}{3}\)

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 \(a=b=c=d\)

b)\(\frac{ac}{bd}=\frac{bkdk}{bd}=k.k=k^2\)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)

=> \(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Đặt k ( với k khác 0 , thuộc Z ) sao cho \(\frac{a}{b}=\frac{c}{d}=k\) => \(a=kb\)  /  \(c=dk\) .

a) Thế vào \(\frac{5a-b}{3a+2b}\) , ta có \(\frac{5kb-3b}{3kb+2b}\)\(=\frac{b\left(5k-3\right)}{b\left(3k+2\right)}\)\(=\frac{5k-3}{3k+2}\)  /  \(\frac{5c-3d}{3c+2d}=\frac{5dk-3d}{3dk-2d}=\frac{d\left(5k-3\right)}{d\left(3k+2\right)}=\frac{\left(5k+3\right)}{\left(3k+2\right)}\)

=> VT = VP

1.

\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)

\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)

Dấu "=" khi \(a=b=c\)

2.

\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)

Dấu "=" khi \(a=b=c=d\)

Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,

Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana

Help me!

Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.