đề sai

day la hoc ki may vay bannn

\(404=3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\ge\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-\dfrac{2}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\le1212\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le2\sqrt{303}\)

Ta có:

\(5a^2+2ab+2b^2=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow P\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{2}{c}+\dfrac{1}{a}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{2\sqrt{303}}{3}\)

Đặt 2a/5b=5b/6c=6c/7d=7d/2a=k

=> k^4=2a/5b.5b/6c.6c/7d.7d/2a=1

=>k=1 hoặc k=-1

Với k=1 thì B=4

Với k=-1 thì B=-4

Vậy B=4 hoặc B=-4

Ta có: 2a = 3b => \(\dfrac{a}{3}=\dfrac{b}{2}\)

Ta có: 5b = 6c => \(\dfrac{b}{6}=\dfrac{c}{5}\)

Ta có: \(\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{6}=\dfrac{c}{5}\Rightarrow\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}\)

và a + 3b - 2c = -5

Áp dụng t/c dãy tỉ số = nhau; ta có:

\(\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+3b-2c}{9+3.6-2.5}=\dfrac{-5}{17}\)

\(\dfrac{a}{9}=\dfrac{-5}{17}\) => a = -45/17

\(\dfrac{b}{6}=\dfrac{-5}{17}\) => b = -30/17

\(\dfrac{c}{5}=\dfrac{-5}{17}\) => c = -25/17

Vậy... a = -45/17

b = -30/17

c = -25/17.

Ta có:

+) \(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{18}=\dfrac{b}{12}\)

+) \(5b=6c\Rightarrow\dfrac{b}{6}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{10}\)

=> \(\dfrac{a}{18}=\dfrac{b}{12}=\dfrac{c}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{18}=\dfrac{b}{12}=\dfrac{c}{10}\Rightarrow\dfrac{a}{18}+\dfrac{3b}{36}-\dfrac{2c}{20}=\dfrac{a+3b-2c}{18+36-20}=-\dfrac{5}{34}\)

Suy ra:

\(\dfrac{a}{18}=-\dfrac{5}{34}\Rightarrow a=-\dfrac{45}{17}\)

\(\dfrac{b}{12}=-\dfrac{5}{34}\Rightarrow b=-\dfrac{30}{7}\)

\(\dfrac{c}{10}=-\dfrac{5}{34}\Rightarrow c=-\dfrac{25}{17}\)

Cách nhanh nhất :

Theo bài ra :

2a=3b=>2a-3b=0 (1)

5b=6c=> 5b-6c=0 (2)

a+3b-2c=-5 (3)

Từ (1) , (2) , (3) ta có hệ :

\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)

\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

thay 28 vao pt nhan tu roi am-gm cho cai do luon

Ps: tim Min

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)

BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)

Ta có:

\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)

\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)

Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)

\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)

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