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\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)
\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)
\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)
\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)
Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)
\(yz\le\frac{\left(y+z\right)^2}{4}\Rightarrow\frac{x^2\left(y+z\right)}{yz}\ge\frac{4x^2}{y+z}\)
Do đó \(P\ge\frac{4x^2}{y+z}+\frac{4y^2}{z+x}+\frac{4z^2}{x+y}\ge\frac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}=2\)(Vì x+y+z = 1)
Vậy Min P= 2. Dấu "=" có <=> x = y = z = 1/3.
Với \(a;b>0\) ta luôn có: \(\frac{a^3+b^3}{a^2+b^2}\ge\frac{a+b}{2}\)
Thật vậy, BĐT tương đương:
\(2\left(a^3+b^3\right)\ge\left(a^2+b^2\right)\left(a+b\right)\)
\(\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)
Áp dụng vào bài toán:
\(P=\frac{x^3+y^3}{x^2+y^2}+\frac{y^3+z^3}{y^2+z^2}+\frac{z^3+x^3}{z^2+x^2}\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z\ge6\)
\(\Rightarrow P_{min}=6\) khi \(x=y=z=2\)
ta có:
\(S\ge\frac{x^3}{x^2+y^2+\frac{x^2+y^2}{2}}+\frac{y^3}{y^2+z^2+\frac{y^2+z^2}{2}}+\frac{z^3}{z^2+x^2+\frac{z^2+x^2}{2}}\)
\(\Rightarrow S\ge\frac{2x^3}{3\left(x^2+y^2\right)}+\frac{2y^3}{3\left(y^2+z^2\right)}+\frac{2z^3}{3\left(z^2+x^2\right)}\Rightarrow\frac{3}{2}S\ge P=\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\)
\(\Rightarrow P=x-\frac{xy^2}{x^2+y^2}+y-\frac{yz^2}{y^2+z^2}+z-\frac{zx^2}{z^2+x^2}\ge\left(x+y+z\right)-\left(\frac{xy^2}{2xy}+\frac{yz^2}{2yz}+\frac{zx^2}{2xz}\right)\)
\(=\left(x+y+z\right)-\frac{1}{2}\left(x+y+z\right)=\frac{9}{2}\)
\(\Rightarrow\frac{3}{2}S\ge\frac{9}{2}\Rightarrow S\ge3\)
Vậy Min S=3 khi x=y=z=3
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Ta có:
\(A=\left(x^2+\frac{1}{8x}+\frac{1}{8x}\right)+\left(y^2+\frac{1}{8y}+\frac{1}{8y}\right)+\left(z^2+\frac{1}{8z}+\frac{1}{8z}\right)+\frac{6}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\ge3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}+3\sqrt[3]{y^2.\frac{1}{8y}.\frac{1}{8y}}+3\sqrt[3]{z^2.\frac{1}{8z}.\frac{1}{8z}}+\frac{6}{8}\frac{9}{x+y+z}\)
\(=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{6}{8}.\frac{9}{\frac{3}{2}}=\frac{27}{4}\)
Dấu "=" xảy ra <=> x = y = z = 1/2
Vậy min A = 27/4 tại x = y = z = 1/2
Áp dụng bđt \(\frac{a^2}{m}+\frac{b^2}{n}+\frac{c^2}{p}\ge\frac{\left(a+b+c\right)^2}{m+n+p}\) được
\(G\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=\frac{2}{2}=1\)
\(G\ge1\Rightarrow MinG=1\Leftrightarrow\hept{\begin{cases}x=y=z>0\\x+y+z=2\end{cases}\Leftrightarrow}x=y=z=\frac{2}{3}\)