Question

cho a,b,c là các số thực dương thỏa ab+bc+ca=1.cmr

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

Answer 1

\(1+a^2=a^2+ab+ac+bc=\left(a+b\right)\left(a+c\right)\)

Tương tự với 2 biểu thức còn lại

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

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

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

\(=b+c-a^2b-a^2c+a+c-ab^2-b^2c\)

\(=2c+a+b-a\left(1-bc-ac\right)-a^2c-b\left(1-bc-ac\right)-b^2c\)

\(=2c+a+b-a+abc+a^2c-a^2c-b+b^2c+abc-b^2c\)

\(=2c+2abc=2c\left(1+ab\right)\)