Với \(a;b;c\ne0\) ta luôn có:
\(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2\ge0\)
\(\Leftrightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}\ge0\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
Dấu "=" xảy ra khi \(a=b=c\ne0\)
Do \(x;y\in N\) *\(\Rightarrow x+y\ge2\)
\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\frac{x+y}{x^2+y^2}\le\frac{2\left(x+y\right)}{\left(x+y\right)^2}=\frac{2}{x+y}\)
\(\Rightarrow\frac{2}{x+y}\ge\frac{3}{5}\Rightarrow x+y\le\frac{10}{3}\)
\(\Rightarrow x+y=\left\{2;3\right\}\)
TH1: \(x=y=1\Rightarrow\frac{x+y}{x^2+y^2}=1\left(ktm\right)\)
TH2: \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
\(\Rightarrow\frac{x+y}{x^2+y^2}=\frac{3}{5}\left(tm\right)\)
\(a+b+c=6abc\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=6\)
\(P=\frac{y^3}{x}+\frac{z^3}{y}+\frac{x^3}{z}=\frac{y^4}{xy}+\frac{z^4}{yz}+\frac{x^4}{zx}\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+yz+zx}\)
\(P\ge\frac{\left(xy+yz+zx\right)^2}{xy+yz+zx}=xy+yz+zx=6\)
\(P_{min}=6\) khi \(x=y=z=\sqrt{2}\Rightarrow a=b=c=\frac{1}{\sqrt{2}}\)
