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Các bạn giải nhanh giúp mình

ÁP dụng BĐT AM-Gm  ta có: 

\(Σ\frac{a^2}{\left(ab+2\right)\left(2ab+1\right)}\ge\frac{4}{9}\cdotΣ\frac{a^2}{\left(ab+1\right)^2}\)

ĐẶt \(a=\frac{x}{y};b=\frac{y}{z};c=\frac{z}{x}\) thì cần cm

\(Σ\frac{a^2}{\left(ab+1\right)^2}=Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{3}{4}\)

\(Σ\left(\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\left(\frac{xz}{y\left(x+z\right)}\right)^2\)

Theo C-S \(Σ\frac{xz}{y\left(x+z\right)}=\frac{\left(xz\right)^2}{xyz\left(x+z\right)}\ge\frac{\left(Σxy\right)^2}{2xy\left(Σx\right)}\ge\frac{3}{2}\)

\(\frac{1}{3}\cdot\left(Σ\frac{xz}{y\left(x+z\right)}\right)^2\ge\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}\)

Đúng hay ta có ĐPCM xyar ra khi a=b=c=1

\(A=\frac{1}{x-\sqrt{x}+1}=\frac{1}{x-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}+\frac{3}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\)

Vì\(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\Rightarrow A=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{4}{3}\)

Dấu "=" xảy ra khi \(\left(\sqrt{x}-\frac{1}{2}\right)^2=0\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)