Đề bài có vấn đề, thay \(a=b=c\) hai vế cho kết quả khác nhau
Ta sẽ chứng minh BĐT mạnh hơn sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+3\sqrt{\dfrac{3\left(a+b+c\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(ab+bc+ca\right)^2}}=3+3\sqrt{Q}\)
Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)
\(\Rightarrow Q\le\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{abc}-3\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow\dfrac{a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)}{abc}\ge3\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}}\)
\(\Leftrightarrow a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\ge\dfrac{3}{\sqrt{2}}\sqrt{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(VT\ge3\sqrt[3]{abc\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\)
Do đó ta chỉ cần chứng minh:
\(8\left(abc\right)^2\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)^2\right]^2\ge\left(abc\right)^3\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow8\left[\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\right]^2\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\dfrac{1}{8}\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^4\ge abc\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) (hiển nhiên đúng theo AM-GM)
