từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)

ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)

=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)

\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )

^_^

ta có \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}=\sqrt{\left(1+a\right)\left(a^2-a+1\right)}.\sqrt{\left(1+b\right)\left(b^2-b+1\right)}\)

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

Tương tự thì \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}\le\dfrac{\left(a^2+2\right)\left(b^2+2\right)}{4}\Rightarrow\dfrac{a^2}{\sqrt{\left(1+a^3\right)\left(1+B^3\right)}}\ge\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)

=\(\dfrac{4a^2\left(c^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)

Tương tự rồi + vào, ta có

...\(\ge4\dfrac{a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)

ta cần chứng minh \(3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)

đến đây nhân tung ra và dùng cô-si tiếp