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

TT...

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

\(\ge6-\frac{y^2\left(x+1\right)}{2y}-\frac{z^2\left(y+1\right)}{2z}-\frac{x^2\left(z+1\right)}{2x}=6-\frac{xy+yz+xz+x+y+z}{2}\)

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

Vậy GTNN của Q là 3 khi x = y = z = 1

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

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

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\) thì \(a^2+b^2+c^2=1\) Ta cần chứng minh:

\(P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)

\(=\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}\)

\(=\frac{a^2}{a\left(1-a^2\right)}+\frac{b^2}{b\left(1-b^2\right)}+\frac{c^2}{c\left(1-c^2\right)}\)

Theo đánh giá bởi AM - GM ta có:

\(a^2\left(1-a^2\right)^2=\frac{1}{2}\cdot2a^2\cdot\left(1-a^2\right)\left(1-a^2\right)\)

\(\le\frac{1}{2}\left(\frac{2a^2+1-a^2+1-a^2}{3}\right)^3=\frac{4}{27}\)

\(\Rightarrow a\left(1-a^2\right)^2\le\frac{2}{3\sqrt{3}}\Leftrightarrow\frac{a^2}{a\left(1-a\right)^2}\ge\frac{3\sqrt{3}}{2}a^2\)

Tương tự rồi cộng lại ta có ngay điều phải chứng minh

Câu 1:

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)

Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)

\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)

Dấu = xảy ra khi x=y=1/2

Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)

CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn

\(Q=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=\frac{1^2}{xy}+\frac{1^2}{yz}+\frac{1^2}{xz}\ge\frac{\left(1+1+1\right)^2}{xy+yz+xz}\)

\(=\frac{9}{xy+yz+zx}\ge\frac{9}{x^2+y^2+z^2}\ge\frac{9}{6}=\frac{3}{2}\).

Dấu " = " xảy ra <=> x = y =z = \(\sqrt{2}\).

Áp dụng bđt AM-GM ta được:

\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)

\(\frac{y^2}{z+x}+\frac{z+x}{4}\ge2\sqrt{\frac{y^2}{z+x}.\frac{z+x}{4}}=y\)

\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}=z\)

Cộng từng vế các bất đẳng thức trên ta được

\(A+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow A\ge\frac{x+y+z}{2}=1\)

Dấu"="xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)

Cách 2:Dù dài hơn Lê Tài Bảo Châu

\(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\left(x+y+z\right)\cdot\frac{x}{y+z}\)

\(\frac{y^2}{z+x}+y=\left(x+y+z\right)\cdot\frac{y}{z+x};\frac{z^2}{x+y}+z=\left(x+y+z\right)\cdot\frac{z}{x+y}\)

Suy ra \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+\left(x+y+z\right)=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

Đến đây thay x+y+z=2 và BĐT netbitt là ra ( chứng minh netbitt nha )

Cách 3:

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

Dấu "=" xảy ra tại \(a=b=c=\frac{2}{3}\)

Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1