Áp dụng BĐT Cauchy-Schwarz ta có:
\(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}=\sqrt{a}\cdot\sqrt{3a+b}+\sqrt{b}\cdot\sqrt{3b+a}\)
\(\le\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}=2\left(a+b\right)\)
\(\Rightarrow\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
Xảy ra khi \(a=b\)
Thắng Nguyễn ơi, bài này dùng cô si được ko bạn
Áp dụng bất đăng thức Bunhiacopxki ta có
\(\sqrt{a\left(3a+b\right)}\)+\(\sqrt{b\left(3b+a\right)}\)=\(\sqrt{a}\)\(\sqrt{3a+b}\)+\(\sqrt{b}\)\(\sqrt{3b+a}\)<(=)\(\sqrt{\left(a+b\right)\left(3a+b+3b+a\right)}\)=2(a+b)suy ra \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)>(=)\(\frac{a+b}{2\left(a+b\right)}\)=\(\frac{1}{2}\)Dấu đẳng thức xảy ra khi và chỉ khi a=b
Ta có : \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}=\frac{2\left(a+b\right)}{2\left(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\right)}=\frac{2\left(a+b\right)}{2\sqrt{a\left(3a+b\right)}+2\sqrt{b\left(3b+a\right)}}=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\)
Áp dụng bất đăng thức côsi ta có :
\(\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)
\(\sqrt{4b\left(3b+a\right)}\le\frac{4b+3b+a}{2}=\frac{7b+a}{2}\)
Cộng vế theo vế ta có : \(\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}\le\frac{7a+b}{2}+\frac{7b+a}{2}=\frac{8a+8b}{2}=4\left(a+b\right)\)
\(\Rightarrow\frac{2\left(a+b\right)}{4\left(a+b\right)}\ge\frac{1}{2}\)( đpcm )
