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Dự đoán khi \(x=y=z=\sqrt{3}\) vậy dc GTNN là \(\frac{3\sqrt{3}}{2}\), cần c/m: \(P\ge\frac{3\sqrt{3}}{2}\)
\(\LeftrightarrowΣ\frac{y^2z^2}{x\left(y^2+z^2\right)}\ge\frac{3}{2}\sqrt{\frac{3}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}}\)
\(\LeftrightarrowΣ\frac{y^3z^3}{y^2+z^2}\ge\frac{3}{2}\sqrt{\frac{3x^4y^4z^4}{x^2y^2+x^2z^2+y^2z^2}}\).Đặt \(\hept{\begin{cases}yz=a\\xz=b\\xy=c\end{cases}}\)
Khi đó ta cần chứng minh \(Σ\frac{a^3}{\frac{ac}{b}+\frac{ab}{c}}\ge\frac{3}{2}\sqrt{\frac{3a^2b^2c^2}{a^2+b^2+c^2}}\)
\(\LeftrightarrowΣ\frac{a^2}{b^2+c^2}\ge\frac{3}{2}\sqrt{\frac{3}{a^2+b^2+c^2}}\) và từ BĐT thuần nhất cuối , ta có thế khẳng định rằng \(a^2+b^2+c^2=3\)
Có nghĩa là ta cần c/m \(Σ\frac{a}{3-a^2}\ge\frac{3}{2}\LeftrightarrowΣ\left(\frac{a}{3-a^2}-\frac{1}{2}\right)\ge0\)
\(\LeftrightarrowΣ\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}\ge0\)\(\LeftrightarrowΣ\left(\frac{\left(a-1\right)\left(a+3\right)}{3-a^2}-\left(a^2-1\right)\right)\ge0\)
\(\LeftrightarrowΣ\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}\ge0\) . XOng!
\(P=\frac{y^2z^2}{x\left(y^2+z^2\right)}+\frac{z^2x^2}{y\left(x^2+z^2\right)}+\frac{x^2y^2}{z\left(x^2+y^2\right)}\)
\(=\frac{1}{x\left(\frac{1}{y^2}+\frac{1}{z^2}\right)}+\frac{1}{y\left(\frac{1}{z^2}+\frac{1}{x^2}\right)}+\frac{1}{z\left(\frac{1}{x^2}+\frac{1}{y^2}\right)}\)
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
Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)
Ta có \(a,b,c>0;a^2+b^2+c^2=1\)
và \(P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)
\(=\frac{a^2}{a\left(1-a^2\right)}+\frac{b^2}{b\left(1-b^2\right)}+\frac{c^2}{c\left(1-c^2\right)}\)
Áp dụng bất đẳng thức Cô-si cho 3 số dương ta có
\(a^2\left(1-a^2\right)^2=\frac{1}{2}.2a^2.\left(1-a^2\right)\left(1-a^2\right)\)
\(\le\frac{1}{2}\left(\frac{2a^2+1-a^2+1-a^2}{3}\right)^3=\frac{4}{27}\)
\(\Rightarrow a\left(1-a^2\right)\le\frac{2}{3\sqrt{3}}\Rightarrow\frac{a^2}{a\left(1-a^2\right)}\ge\frac{3\sqrt{3}}{2}a^2\)(1)
Tương tự \(\frac{b^2}{b\left(1-b^2\right)}\ge\frac{3\sqrt{3}}{2}b^2\)(2)
\(\frac{c^2}{c\left(1-c^2\right)}\ge\frac{3\sqrt{3}}{2}c^2\)(3)
từ (1),(2) và (3) ta có \(P\ge\frac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\frac{3\sqrt{3}}{2}\)
Vậy Min của \(P=\frac{3\sqrt{3}}{2}\)Khi x=y=z\(=\sqrt{3}\)
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
Lời giải:
\(\frac{1}{x^2}=1-\frac{1}{y^2}-\frac{1}{z^2}<1\Rightarrow x^2-1>0\)
\(P=\frac{y^2z^2}{x(y^2+z^2)}+\frac{x^2z^2}{y(x^2+z^2)}+\frac{x^2y^2}{z(x^2+y^2)}\)
\(=\frac{1}{x(\frac{1}{y^2}+\frac{1}{z^2})}+\frac{1}{y(\frac{1}{x^2}+\frac{1}{z^2})}+\frac{1}{z(\frac{1}{x^2}+\frac{1}{y^2})}\)
\(=\frac{1}{x(1-\frac{1}{x^2})}+\frac{1}{y(1-\frac{1}{y^2})}+\frac{1}{z(1-\frac{1}{z^2})}\)
\(=\frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1}\)
Xét đánh giá sau:
\(\frac{x}{x^2-1}-\frac{3\sqrt{3}}{2x^2}=\frac{(x-\sqrt{3})^2(2x+\sqrt{3})}{2x^2(x^2-1)}\geq 0, \forall x^2>1\)
\(\Rightarrow \frac{x}{x^2-1}\geq \frac{3\sqrt{3}}{2x^2}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow P=\frac{x}{x^2-1}+\frac{y}{y^2-1}+\frac{z}{z^2-1}\geq \frac{3\sqrt{3}}{2}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{3\sqrt{3}}{2}\)
Vậy \(P_{\min}=\frac{3\sqrt{3}}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
SOS get it <(")
\(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)->\left(a;;bc\right)\text{for}\left(a;b;c>0\text{and}a^2+b^2+c^2=1\right)\)
\(\text{Khido}P=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)
\(\text{Ta se cm}\sum_{cyc}\frac{a}{b^2+c^2}\ge\frac{3\sqrt{3}}{2}\)\(\text{Viet lai BDT can chung minh}\)
\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}}{2\sqrt{x^2+y^2+z^2}}\)
\(\text{Chuan hoa}a^2+b^2+c^2=3\text{ta can cm:}\)
\(\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{3-a^2}-\frac{1}{2}+\frac{b}{3-b^2}-\frac{1}{2}+\frac{c}{3-c^2}-\frac{1}{2}\ge0\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{a}{3-a^2}-\frac{1}{2}-\frac{1}{2}\left(x^2-1\right)\right)\ge0\)
\(\Leftrightarrow\frac{a\left(a+2\right)\left(a-1\right)^2}{3-a^2}+\frac{b\left(b+2\right)\left(b-1\right)^2}{3-b^2}+\frac{c\left(c+2\right)\left(c-1\right)^2}{3-c^2}\ge0\)
kho ghe
\(a+b+c=1\)
\(P=\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2}\)