Question

[Tiểu Bảo Bảo]

Cho các số dương a,b,c thỏa mãn \(\left\{{}\begin{matrix}a+b+c=2\\a^2+b^2+c^2=2\end{matrix}\right.\)

Chứng minh rằng: \(a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}+b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\dfrac{\left(1+a^2\right)\left(1+b^2\right)}{1+c^2}}=2\)

Answer 1

[Phương Ann]

• Vì a, b, c đều dương và a + b + c = 2

nên \(0< a,b,c< 2\)

• Theo gt, ta có:

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

\(\Rightarrow\left(2-a\right)^2-2+a^2=2bc\)

\(\Rightarrow bc=\dfrac{\left(4-4a+a^2\right)-2+a^2}{2}=\dfrac{2a^2-4a+2}{2}=\left(a-1\right)^2\)

\(\Rightarrow b^2c^2=\left(a-1\right)^4\)

• Ta lại có: \(a\sqrt{\dfrac{\left(1+b^2\right)\left(1+c^2\right)}{1+a^2}}=a\sqrt{\dfrac{1+b^2+c^2+b^2c^2}{1+a^2}}\)

\(=a\sqrt{\dfrac{3-a^2+\left(a-1\right)^4}{1+a^2}}=a\sqrt{\dfrac{a^4-4a^3+5a^2-4a-4}{1+a^2}}\)

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

• Tương tự, ta cũng có: \(b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}=b\left(2-b\right)\)

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

• Suy ra \(a\sqrt{\dfrac{\left(1+a^2\right)\left(a-2\right)^2}{1+a^2}}+b\sqrt{\dfrac{\left(1+a^2\right)\left(1+c^2\right)}{1+b^2}}+c\sqrt{\dfrac{\left(1+b^2\right)\left(1+a^2\right)}{1+c^2}}\)

\(=2\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(đpcm\right)\)