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Có \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\ge9\)
\(\Leftrightarrow\dfrac{a+1}{a}.\dfrac{b+1}{b}\ge9\)
\(\Leftrightarrow ab+a+b+1\ge9ab\) ( vì ab >0)
\(\Leftrightarrow a+b+1\ge8ab\)
\(\Leftrightarrow2\ge8ab\) \(\left(a+b=1\right)\)
\(\Leftrightarrow1\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\left(a+b=1\right)\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng)
\(\Leftrightarrowđpcm\)
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Ta co:\(1\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le\frac{1}{4}\)
Dat \(P=a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)
\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)
\(=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+\frac{15}{16}.\frac{a^2+b^2}{a^2b^2}\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{16}.\frac{2}{ab}\ge1+\frac{15}{16}.\frac{2}{\frac{1}{4}}=\frac{17}{2}\)
Dau '=' xay ra \(a=b=\frac{1}{2}\)
Vay \(P_{min}=\frac{17}{2}\)khi \(a=b=\frac{1}{2}\)
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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0\)
\(\left(a+b+c\right)^2=1\Leftrightarrow a^2+b^2+c^2+2.\left(ab+bc+ca\right)=1\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=1\)
\(\Leftrightarrow a^2+b^2+c^2=1\)
Đề sai rồi bạn ơi!