Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:

\(\left\{{}\begin{matrix}\dfrac{a+b}{2}\ge\sqrt{ab}\\\dfrac{b+c}{2}\ge\sqrt{bc}\\\dfrac{a+c}{2}\ge\sqrt{ac}\end{matrix}\right.\)

Cộng theo 3 vế ta có:

\(\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{a+c}{2}\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\Rightarrow\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}b+\dfrac{1}{2}c+\dfrac{1}{2}a+\dfrac{1}{2}c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\left(đpcm\right)\)

\(a=b=c\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+c^2=2bc\\a^2+c^2=2ac\end{matrix}\right.\)

Cộng theo 3 vế ta có:

\(a^2+b^2+b^2+c^2+a^2+c^2=2ab+2bc+2ac\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)

Ngược lại,khi \(a\ne b\ne c\) thì \(\left\{{}\begin{matrix}a^2+b^2>2ab\\b^2+c^2>2bc\\a^2+c^2>2ac\end{matrix}\right.\) ta có thể dễ dàng cm được \(a^2+b^2+c^2>ab+bc+ac\)

\(VT=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{2}\)

Tượng tự ta có \(\hept{\begin{cases}\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{a+b}}{2}\end{cases}}\)

\(\Rightarrow VT\le\frac{\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{c}{a+c}+\frac{a}{c+a}\right)+\left(\frac{c}{b+c}+\frac{b}{c+b}\right)}{2}\)

\(\Rightarrow VT\le\frac{\frac{a+b}{a+b}+\frac{c+a}{c+a}+\frac{b+c}{b+c}}{2}=\frac{3}{2}\) ( đpcm ) 

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)

cauchy - schwarz là bđt Cauchy à bạn

hihih ~ e mới lớp 8 ~ năm sau nha

1. \(\sqrt[3]{8}=2.\)

2. \(A=\sqrt{16a^2}=4\left|a\right|\)

\(\Rightarrow\left[{}\begin{matrix}A=4a\left(a\ge0\right)\\A=-4a\left(a< 0\right)\end{matrix}\right..\)

3. \(B=\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}=\dfrac{23\sqrt{6}}{46}=\dfrac{\sqrt{6}}{2}.\)

4. C.

lớp 9 làm quen không bạn ^^

a)Áp dụng BĐT AM-GM ta có

\(\frac{ab\sqrt{ab}}{a+b}\le\frac{ab\sqrt{ab}}{2\sqrt{ab}}=\frac{ab}{2}\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\frac{bc\sqrt{bc}}{b+c}\le\frac{bc}{2};\frac{ac\sqrt{ac}}{a+c}\le\frac{ac}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=Σ\frac{ab\sqrt{ab}}{a+b}\le\frac{ab+bc+ca}{2}=VP\)

Khi \(a=b=c\)

b)Áp dụng tiếp AM-GM:

\(b\sqrt{a-1}\le\frac{b\left(a-1+1\right)}{2}=\frac{ab}{2}\)

\(a\sqrt{b-1}\le\frac{a\left(b-1+1\right)}{2}=\frac{ab}{2}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=b\sqrt{a-1}+a\sqrt{b-1}\le ab=VP\)

Khi \(a=b=1\)