\(\frac{x+1}{1+y^2}=\frac{\left(x+1\right)\left(y^2+1\right)-y^2\left(x+1\right)}{1+y^2}=x+1-\frac{y^2\left(x+1\right)}{1+y^2}\ge x+1-\frac{xy+y}{2}\)

Tương tự ta có:

\(\frac{y+1}{z^2+1}\ge y+1-\frac{yz+z}{2}\)

\(\frac{z+1}{1+x^2}\ge z+1-\frac{zx+x}{2}\)

Cộng vế theo vế ta có:

\(Q\ge3+\left(x+y+z\right)-\frac{x+y+z+xy+yz+zx}{2}\)

\(=3+\frac{x+y+z-xy-yz-zx}{2}\)

Có BĐT phụ sau:

\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) ( tự cm )

\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)

Khi đó \(P\ge3\)

Dấu "=" xảy ra tại \(x=y=z=1\)

Áp dụng BĐT Cauchy cho 3 số dương, ta được:

\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)

\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)

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

Do \(x;y;z>0\) và \(x^2+y^2+z^2=3\)

Nên \(0< x;y;z< \sqrt{3}\)

Ta có: \(\frac{1}{x+y+z}\le\frac{1}{9x}+\frac{1}{9y}+\frac{1}{9z}\)

\(\Rightarrow A\ge x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}-\frac{1}{9x}-\frac{1}{9y}-\frac{1}{9z}\)

\(\Leftrightarrow A\ge x+\frac{8}{9x}+y+\frac{8}{9y}+z+\frac{8}{9z}\)

Ta chứng minh: \(x+\frac{8}{9x}\ge\frac{x^2+33}{18}\)

\(\Leftrightarrow\left(x-1\right)^2\left(16-x\right)\ge\)

Do đó \(A\ge\frac{x^2+y^2+z^2+99}{18}=\frac{102}{18}=\frac{17}{3}\)

Dấu = xảy ra khi x=y=z=1

Dòng thứ 3 từ dưới lên là \(\left(x-1\right)^2\left(16-x\right)\ge0\)

                              Đúng do \(0< x< \sqrt{3}< 16\)

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)

Ta có: \(\frac{x+1}{y^2+1}=\left(x+1\right).\frac{1}{y^2+1}=\left(x+1\right)\left(1-\frac{y^2}{y^2+1}\right)\)

\(\ge\left(x+1\right)\left(1-\frac{y^2}{2y}\right)=x+1-\frac{y\left(x+1\right)}{2}\)

Thiết lập hai BĐT còn lại tương tự và cộng theo vế:

\(P\ge\left(x+y+z+3\right)-\frac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)

\(=6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\) (*)

Lại có BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

Thật vậy,ta có: BĐT \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)

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

Thay vào (*),ta có: \(P\ge6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\)

\(\ge6-\frac{\frac{\left(x+y+z\right)^2}{3}+3}{2}=6-\frac{3+3}{2}=3\)

Dấu "=" xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=3\Leftrightarrow x=y=z=1\)

Bài t đúng 100% nhá,đứa nào tk sai t nhở? ngon vô làm lại=)

\(taco:\)

\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}\)

\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{2}\ge3\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=\frac{3}{2}\)

\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge3\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=\frac{3}{2}\)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)

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

^^

Mình giải lại bài này cho đầy đủ hơn nhé: (nãy chỉ là hướng dẫn thôi)

Ta sẽ c/m: \(\frac{1}{x^2+x}\ge-\frac{3}{4}x+\frac{5}{4}\) (1).Thật vậy,xét hiệu hai vế,ta có:

\(VT-VP=\frac{\left(3x+4\right)\left(x-1\right)^2}{4\left(x^2+x\right)}\ge0\)

Suy ra \(VT\ge VP\).Vậy (1) đúng.

Thiết lập hai BĐT còn lại tương tự và cộng theo vế,ta có:

\(VT\ge-\frac{3}{4}\left(x+y+z\right)+\frac{5}{4}.3=\frac{3}{2}^{\left(đpcm\right)}\)

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