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Ta có:\(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a\left(a+b\right)+c\left(a+b\right)}}\)
\(=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) (Áp dụng BĐT AM-GM)
Tương tự với hai BĐT còn lại và cộng theo vế ta thu được đpcm.
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}\le\sqrt{2\left(\frac{2}{a}+\frac{2}{b}\right)}=2\sqrt{\frac{a+b}{ab}}\)
Tương tự: \(\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le2\sqrt{\frac{b+c}{bc}}\) ; \(\sqrt{\frac{2}{c}}+\sqrt{\frac{2}{a}}\le2\sqrt{\frac{c+a}{ca}}\)
Cộng vế với vế ta sẽ có điều phải chứng minh
Không làm mất tính tổng quát của bài toán, giả sử \(a\ge b\ge c\)(1)
Có \(\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}=\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
Từ (1) => \(\hept{\begin{cases}\frac{2}{a}\le\frac{1}{a}+\frac{1}{b}\\\frac{2}{b}\le\frac{1}{b}+\frac{1}{c}\\\frac{2}{c}\le\frac{1}{a}+\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{\frac{2}{a}}\le\sqrt{\frac{1}{a}+\frac{1}{b}}\\\sqrt{\frac{2}{b}}\le\sqrt{\frac{1}{b}+\frac{1}{c}}\\\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{a}+\frac{1}{c}}\end{cases}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{1}{b}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{a}}+\sqrt{\frac{1}{c}+\frac{1}{b}}\)
=>\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)
Ta có đpcm
ta có:
\(c+ab=c.1+ab=c\left(a+b+c\right)+ab=ca+cb+c^2+ab=\left(c+a\right)\left(c+b\right)\)
tương tự như vậy thì \(P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)
áp dụng bđt cô si ta có:
\(\frac{a}{a+c}+\frac{b}{b+c}\ge2\sqrt{\frac{ab}{\left(c+a\right)\left(b+c\right)}};\frac{b}{a+b}+\frac{c}{c+a}\ge2\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}};\frac{a}{a+b}+\frac{c}{b+c}\ge2\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+a}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)=\frac{3}{2}\left(Q.E.D\right)\)
b) Ta có:
\(\frac{a}{\sqrt{b^2+3}}+\frac{a}{\sqrt{b^2+3}}+\frac{b^2+3}{8}+\frac{a^2}{2}\)\(\ge\)\(4\sqrt[4]{\frac{a^4}{16}}=2a\)
\(\frac{b}{\sqrt{c^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c^2+3}{8}+\frac{b^2}{2}\ge4\sqrt[4]{\frac{b^4}{16}}=2b\)
\(\frac{c}{\sqrt{a^2+3}}+\frac{c}{\sqrt{a^2+3}}+\frac{a^2+3}{8}+\frac{c^2}{2}\ge4\sqrt[4]{\frac{c^4}{16}}=2c\)
Cộng lại ta đươc:
\(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)+\)\(\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)\(\ge2\left(a+b+c\right)\)
⇒ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5\left(a^2+b^2+c^2\right)+9}{8}\)(1)
Lại có: \(a^2+1\ge2a\); \(b^2+1\ge2b\); \(c^2+1\ge2c\)
Suy ra \(a^2+b^2+c^2\ge2\left(a+b+c\right)-3=3\)
Khi đó (1)⇔ \(2\left(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\right)\ge\)\(6-\frac{5.3+9}{8}=3\)
⇒ \(\frac{a}{\sqrt{b^2+3}}+\frac{b}{\sqrt{c^2+3}}+\frac{c}{\sqrt{a^2+3}}\ge\frac{3}{2}\)
Dấu "=" xảy ra ⇔ \(a=b=c=1\)
\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\Rightarrow\sqrt{a^2+3b^2}\ge\frac{a+3b}{2}\)
\(\Rightarrow P=\sum\frac{ab}{\sqrt{a^2+3b^2}}\le2\sum\frac{ab}{a+3b}=2\sum\frac{ab}{a+b+b+b}\)
\(\Rightarrow P\le\frac{1}{8}\sum ab\left(\frac{1}{a}+\frac{3}{b}\right)=\frac{1}{8}\sum\left(3a+b\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
"=" \(\Leftrightarrow a=b=c=1\)
