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

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

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

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 \(a=b=c=d\)

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.

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:

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

Đẳng thức xảy ra khi a=b=c=d

là \(\frac{2}{3}\) nha

Bài 2:

Áp dụng Bdt Cauchy-Schwarz dạng engel, ta có

\(VT\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\)

Mà theo Bđt cosi 

\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\)

\(=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)

ta cs a/b=c/d=>a/c=b/d

=>2a+3b/2c+3d=3a-4b/3c-4d

=>2a+3b/3a-4b=2c+3d/3c-4d

=>bai toan dc c/m

Cau b tuong tu nha ban

don't forget tick me

a) Ta có \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}\) (1).

\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\) (2).

Từ (1) và (2) \(\Rightarrow\frac{2a+3b}{2c+3d}=\frac{3a-4b}{3c-4d}.\)

\(\Rightarrow\frac{2a+3b}{3a-4b}=\frac{2c+3d}{3c-4d}\left(đpcm\right).\)

Chúc bạn học tốt!

Áp dụng TCDTSBN ta có:

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\) (vì a+b+c+d khác 0)

=>a=b=c=d

=>M=\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{1}{2}\cdot4=2\)

Ta có:a/b=b/c=c/d=d/a

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:a/b=b/c=c/d=(a+b+c+d)/(b+c+d+a)=1

=>a=b=c=d(vì a/b=b/c=c/d=d/a=1)

Thay vào M sau đó tìm được M=2

\(\Leftrightarrow\frac{\sqrt{bc}}{\sqrt{5a\left(3a+2b\right)}}+\frac{\sqrt{ac}}{\sqrt{5b\left(3b+2c\right)}}+\frac{\sqrt{ab}}{\sqrt{5c\left(3c+2a\right)}}\ge\frac{3}{5}\)

\(\Leftrightarrow\frac{bc}{\sqrt{5ab\left(3ac+2bc\right)}}+\frac{ac}{\sqrt{5bc\left(3ab+2ac\right)}}+\frac{ab}{\sqrt{5ac\left(3bc+2ab\right)}}\ge\frac{3}{5}\)

Thật vậy, theo AM-GM ta có:

\(VT\ge\frac{2bc}{5ab+2bc+3ac}+\frac{2ac}{3ab+5bc+2ac}+\frac{2ab}{2ab+3bc+5ac}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\)

\(\Rightarrow VT\ge\frac{2x}{2x+3y+5z}+\frac{2y}{5x+2y+3z}+\frac{2z}{3x+5y+2z}=\frac{2x^2}{2x^2+3xy+5zx}+\frac{2y^2}{5xy+2y^2+3yz}+\frac{2z^2}{3zx+5yz+2z^2}\)

\(\Rightarrow VT\ge\frac{\left(x+y+z\right)^2}{\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+2\left(xy+yz+zx\right)}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)

\(\Rightarrow VT\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{3}{5}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)