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

Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.

Lời giải vắn tắt:

\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)

( thay \(a^2+b^2+c^2=1\))

\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)

\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

\(P=\sqrt{a^2b^2+a^2}+\sqrt{b^2c^2+b^2}+\sqrt{c^2a^2+c^2}\)

\(P\ge\sqrt{\left(ab+bc+ca\right)^2+\left(a+b+c\right)^2}\)

\(P\ge\sqrt{\left(ab+bc+ca\right)^2+3\left(ab+bc+ca\right)}=2\)

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

\(VT=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+c\right)\left(b+a\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\le_{AM-GM}\dfrac{a+b+a+c}{2}+\dfrac{b+c+b+a}{2}+\dfrac{c+a+c+b}{2}=2\left(a+b+c\right)=VP\) (đpcm)

Đầy đủ hơn 1 tí nhé

Theo gt : ab + bc + ca = 1 nên a² + 1 = a² + ab + bc + ca

                                                            = ( a + b )( a + c )

- Áp dụng bđt Cauchy ta có :

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

- Tương tư ta cũng có : 

\(\sqrt{b^2+1}\le\frac{\left(b+a\right)+\left(b+c\right)}{2}\)và \(\sqrt{c^2+1}\le\frac{\left(c+a\right)+\left(c+b\right)}{2}\)

Từ đó suy ra : VT \(\le\frac{\left(a+b\right)+\left(a+c\right)+\left(b+a\right)+\left(b+c\right)+\left(c+a\right)+\left(c+b\right)}{2}\)

                                   \(\le2\left(a+b+c\right)=VP\left(đpcm\right)\)

ta có \(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}.\sqrt{ab+2c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}\sqrt{ab+2c^2}}\)

Áp dụng bất đẳng thức cô si ta có 

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

=\(\frac{1}{2}\left(2ab+a^2+b^2+2c^2\right)=\frac{1}{2}\left[\left(a+b\right)^2+2c^2\right]\le\frac{1}{2}\left(2a^2+2b^2+2c^2\right)=\left(a^2+b^2+c^2\right)\) =1

=> \(\frac{ab+2c^2}{...}\ge\frac{ab+2c^2}{1}=2c^2+ab\)

tương tự + vào thì e sẽ ra điều phải chứng minh

Nhà hàng Tôm hùm kính chào quý khách ĐC : 255 Nguyễn Huệ, Q tân bình , TP HCM

mik ko bt,ahihi