Question

[Phạm Trần Tuyết Ninh]

Cho x, y, z > 0 thoả mãn : x+y+z=3 . Tìm GTNN của : \(P=\frac{x+1}{1+y^2}+\frac{y+1}{1+z^2}+\frac{z+1}{1+x^2}\)

Answer 1

[Y]

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

+ \(\frac{1}{x^2+1}=\frac{x^2+1-x^2}{x^2+1}=1-\frac{x^2}{x^2+1}\)

+ \(x^2+1\ge2x\forall x\)

\(\Rightarrow\frac{x^2}{x^2+1}\le\frac{x^2}{2x}=\frac{x}{2}\)

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

\(\Rightarrow\frac{1}{x^2+1}\ge1-\frac{x}{2}\)

Dấu "=" xảy ra <=> x = 1

+ Tương tự ta cm đc :

\(\frac{1}{y^2+1}\ge1-\frac{y}{2}\). Dấu "=" xảy ra <=> y = 1

\(\frac{1}{z^2+1}\ge1-\frac{z}{2}\). Dấu "=" xảy ra <=> z = 1

Do đó : \(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge3-\left(\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\right)\)

\(\Rightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\ge3-\frac{3}{2}=\frac{3}{2}\) (1)

Dấu "=" xảy ra <=> x = y = z = 1.

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

\(\Rightarrow\frac{x}{y^2+1}\ge x-\frac{xy^2}{2y}=x-\frac{xy}{2}\) ( do \(y^2+1\ge2y\forall y\) )

Dấu "=" xảy ra <=> y = 1.

Tương tự : \(\frac{y}{z^2+1}\ge y-\frac{yz}{2}\). Dấu "=" xảy ra <=> z = 1.

\(\frac{z}{x^2+1}\ge z-\frac{zx}{2}\). Dấu "=" xảy ra <=> x = 1.

Do đó : \(\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge\left(x+y+z\right)-\frac{xy+yz+zx}{2}\)

\(\Rightarrow\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge3-\frac{\frac{\left(x+y+z\right)^2}{3}}{2}\)

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

\(\Rightarrow\frac{x}{y^2+1}+\frac{y}{z^2+1}+\frac{z}{x^2+1}\ge3-\frac{3}{2}=\frac{3}{2}\) (2)

Dấu "=" xảy ra <=> x = y = z = 1.

Từ (1) và (2) suy ra

\(P\ge\frac{3}{2}+\frac{3}{2}=3\)

P = 3 \(\Leftrightarrow x=y=z=1\)

Vậy Min P = 3 \(\Leftrightarrow x=y=z=1\).