Đặt \(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\) = A
Ta có A^2 = \(\left(\sqrt{\left(a-c\right).c}+\sqrt{c.\left(b-c\right)}\right)^2\)
Áp dụng bđt bunhiacopxki ta có A^2 <= \(\left(\sqrt{a-c}^2+\sqrt{c^2}\right).\left(\sqrt{c^2}+\sqrt{b-c^2}\right)\)
= (a-c+c).(c+b-c) = ab
<=> A<= \(\sqrt{ab}\)=> ĐPCM
Dấu"=" <=> a-c = c và c = b-c
<=> a=b=2c>0
Ta có bất đẳng thức bunhicopxki
\(\sqrt{ax}+\sqrt{by}\le\sqrt{\left(a+x\right)\left(b+y\right)}\)
Áp dụng vào ta có:
\(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{\left(a-c+c\right)\left(b-c+c\right)}\le\sqrt{ab}\)
Dấu bằng xảy ra khi a-c = b-c
Đặt: \(\frac{c}{a}=x;\frac{c}{b}=y\left(0< x;y< 1\right)\)
\(BĐT\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)
\(\Leftrightarrow\sqrt{y\left(1-x\right)}+\sqrt{x\left(1-y\right)}\le1\)
Áp dụng BĐT AM-GM, ta có:
\(\sqrt{y\left(1-x\right)}\le\frac{y+1-x}{2}\)
\(\sqrt{x\left(1-y\right)}\le\frac{x+1-y}{2}\)
\(\RightarrowĐPCM\)
