Đề bài sai

Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\Rightarrow xyz=1\)

Đặt vế trái BĐT cần chứng minh là P, ta có:

\(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

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

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{xz\left(xy+y+1\right)}+\dfrac{x}{x\left(yz+z+1\right)}+\dfrac{1}{zx+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x.xyz+xyz+xz}+\dfrac{x}{xyz+xz+1}+\dfrac{1}{xz+x+1}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{xz}{x+1+xz}+\dfrac{x}{1+xz+1}+\dfrac{1}{xz+x+1}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\) \(\Rightarrow xyz=1\)  (x;y;z > 0 do a;b;c>0)

Cần c/m : \(VT=\dfrac{y^2+z^2}{x}+\dfrac{x^2+z^2}{y}+\dfrac{x^2+y^2}{z}\ge x+y+z+3=VP\) 

Dễ dàng c/m : VT \(\ge2\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\)   (1)

Thấy : \(\dfrac{xy}{z}+\dfrac{xz}{y}\ge2x\)  . CMTT : \(\dfrac{xz}{y}+\dfrac{yz}{x}\ge2z;\dfrac{yz}{x}+\dfrac{xy}{z}\ge2y\)

Suy ra : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge x+y+z\)

Có : \(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge3\sqrt[3]{xyz}=3\)

Suy ra : \(2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\right)\ge x+y+z+3\left(2\right)\)

Từ (1) ; (2) suy ra : \(VT\ge VP\)

" = " \(\Leftrightarrow x=y=z=1\Leftrightarrow a=b=c=1\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\) với \(x;y;z>0\Rightarrow xyz=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có: \(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

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

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

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

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{xy}{y+1+xy}+\dfrac{y}{1+xy+y}\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

bài này khó giúp hộ em với

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Lời giải:

Đặt \(\left(\sqrt{\frac{a}{b}},\sqrt{\frac{b}{c}},\sqrt{\frac{c}{a}}\right)=(x,y,z)\). BĐT cần chứng minh chuyển về:

\(x^3+y^3+z^3\geq x^2+y^2+z^2\) với \(xyz=1\)

Thật vậy, áp dụng BĐT Cauchy-Schwarz:

\((x^3+y^3+z^3)(x+y+z)\geq (x^2+y^2+z^2)^2\)

\(\Leftrightarrow x^3+y^3+z^3\geq \frac{(x^2+y^2+z^2)^2}{x+y+z}\)(1)

Theo BĐT AM-GM:

\(x^2+y^2+z^2\geq xy+yz+xz\Leftrightarrow 2(x^2+y^2+z^2)\geq 2(xy+yz+xz)\)

\(\Leftrightarrow 3(x^2+y^2+z^2)\geq (x+y+z)^2\)

\(\Leftrightarrow (x^2+y^2+z^2)\geq \frac{(x+y+z)^2}{3}\geq \frac{(x+y+z).3\sqrt[3]{xyz}}{3}=x+y+z\) (2)

Từ (1),(2)\(\Rightarrow x^3+y^3+z^3\geq x^2+y^2+z^2\) (đpcm)

Dấu bằng xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c\)