Sorry ko bt làm !

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

\(S=\left(1+\dfrac{2a}{3b}\right)\left(1+\dfrac{2b}{3c}\right)\left(1+\dfrac{2c}{3d}\right)\left(1+\dfrac{2d}{3a}\right)\)

có \(1+\dfrac{2a}{3b}\ge2\sqrt{\dfrac{2a}{3b}}\)(BDT AM-GM)

\(=>1+\dfrac{2b}{3c}\ge2\sqrt{\dfrac{2b}{3c}}\)

\(=>1+\dfrac{2c}{3d}\ge2\sqrt{\dfrac{2c}{3d}}\)

\(=>1+\dfrac{2d}{3a}\ge2\sqrt{\dfrac{2d}{3a}}\)

\(=>S\ge16\sqrt{\dfrac{2a.2b.2c.2d}{3a.3b.3c.3d}}=16\sqrt{\dfrac{16abcd}{81abcd}}=16\sqrt{\dfrac{16}{81}}=\dfrac{64}{9}\)

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!

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)

\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)

Do đó: \(\dfrac{3a+2b}{a}=\dfrac{3c+2d}{c}\)

b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)

\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)

Do đó: \(\dfrac{2a-3b}{b}=\dfrac{2c-3d}{d}\)

c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)

\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)

Do đó: \(\dfrac{a}{a-2b}=\dfrac{c}{c-2d}\)