\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\frac{b}{ab}+\frac{a}{ab}\ge\frac{4}{a+b}\)
\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(a+b\right)\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\left(đpcm\right)\)
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) \(\left(ĐK:a>0;b>0\right)\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(a+b\right)\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) (BĐT luôn đúng)
Vậy \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Áp dụng BĐT Cauchy-schwarz ta có:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{2^2}{a+b}=\frac{4}{a+b}\)
đpcm
Tham khảo nhé~
Bỏ bớt mấy cái chứng minh lùng phùng.Ta hãy nhìn sang BĐT Svac cho nhanh!
Ta có BĐT Svac: \(\frac{a^2}{b}+\frac{c^2}{d}\ge\frac{\left(a+c\right)^2}{b+d}\)
Áp dụng vào bài,ta có: \(\frac{1}{a}+\frac{1}{b}=\frac{\left(1+1\right)^2}{a+b}=\frac{4}{a+b}^{\left(đpcm\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{a}=\frac{1}{b}\Leftrightarrow a=b\)