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Từ \(\frac{a^2+b^2}{a-2b}=2\Rightarrow a^2+b^2=2\left(a-2b\right)\)
\(\Leftrightarrow a^2+b^2=2a-4b\)
\(\Leftrightarrow a^2+b^2+4b=2a\)
\(\Leftrightarrow a.a+b.b+4b=2.a\)
\(\Leftrightarrow a.a+b\left(b+4\right)=2.a\)
\(\Leftrightarrow2.a-a.a=b\left(b+4\right)\)
\(\Leftrightarrow\frac{a}{b}=\frac{b+4}{2-a}\)
Mà muốn P lớn nhất thì a,b phải lớn nhất \(\Rightarrow a=b+4;b=2-a\)
\(\Leftrightarrow a+b=2\Leftrightarrow b+4+b=2\Leftrightarrow2b=-2\Rightarrow b=-1;a=3\)
\(\Rightarrow P=8a+4b=24-4=20\)
Ta có: \(b=0,25P-2a\) thế ngược lên trên ta được
\(\frac{a^2+\left(0,25P-2a\right)^2}{a-2\left(0,25P-2a\right)}=2\)
\(\Leftrightarrow80a^2-a\left(16P+160\right)+P^2+16P=0\)
Để PT có nghiệm thì:
\(\Delta'\ge0\)
Làm tiếp nhé
\(M=\frac{\left(a+1\right)^2+2a}{a\left(a+1\right)}+\frac{\left(b+1\right)^2+2b}{b\left(b+1\right)}+\frac{\left(c+1\right)^2+2c}{c\left(c+1\right)}\)
\(M=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+1}{c}+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(M=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(M\ge3+\frac{9}{a+b+c}+2\left(\frac{9}{a+b+c+3}\right)\ge3+3+3=9\)
Dấu "=" xảy ra khi a=b=c=1
Từ \(2a+2b+2c=3abc\)
\(\Leftrightarrow\frac{2}{3bc}+\frac{2}{3ac}+\frac{2}{3ab}=1\left(1\right)\)
Khi đó \(P=\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}\ge3\sqrt[3]{\frac{b}{a^2}\cdot\frac{c}{b^2}\cdot\frac{a}{c^2}}=3\sqrt[3]{\frac{1}{abc}}\)
\(P_{Min}\) xảy ra khi \(\frac{b}{a^2}+\frac{c}{b^2}+\frac{a}{c^2}=3\sqrt[3]{\frac{1}{abc}}\forall a=b=c\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow a=b=c=\sqrt{2}\)
Khi đó \(P_{Min}=3\sqrt[3]{\frac{1}{abc}}-\frac{2}{a^2}-\frac{2}{b^2}-\frac{2}{c^2}=\frac{3\sqrt{2}-6}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\sqrt{2}\)
Bài này giải như này cơ:
\(2a+2b+2c=3abc\)\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{3}{2}\)
\(P=\frac{\left(a-1\right)+\left(b-1\right)}{a^2}+\frac{\left(b-1\right)+\left(c-1\right)}{b^2}+\frac{\left(c-1\right)+\left(a-1\right)}{c^2}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\left(a-1\right)\left(\frac{1}{a^2}+\frac{1}{c^2}\right)+\left(b-1\right)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\left(c-1\right)\left(\frac{1}{b^2}+\frac{1}{c^2}\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\ge\frac{2\left(a-1\right)}{ac}+\frac{2\left(b-1\right)}{ab}+\frac{2\left(c-1\right)}{bc}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3\)
\(\ge\sqrt{3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}-3=\sqrt{3.\frac{3}{2}}-3=\frac{3\sqrt{2}-6}{2}\)
Vậy \(minP=\frac{3\sqrt{2}-6}{2}\Leftrightarrow a=b=c=\sqrt{2}\)