Áp dụng BĐT AM-GM: \(VT\le\sum\dfrac{1}{\sqrt{a^2+1}.\sqrt{2a}.2\sqrt{bc}}=\sum\dfrac{1}{2\sqrt{2}\sqrt{a^2+1}}\)
Ta đi chứng minh \(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
Giả sử c=max{a, b, c}.Suy ra \(c\ge1\) nên \(ab\le1\). Ta có bổ đề:
\(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}\le\dfrac{2}{\sqrt{1+ab}}\)(*)
#cm: Áp dụng Bunyakovsky: \(VT_{(*)} \)\(\le\sqrt{2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\right)}\)
Xét \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}-\dfrac{2}{ab+1}=\dfrac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\le0\)
Nên \(VT_{(*)}\)\(\le\sqrt{2.\dfrac{2}{ab+1}}=\dfrac{2}{\sqrt{ab+1}}\), suy ra đpcm.
Do đó \(VT\le\dfrac{2}{\sqrt{ab+1}}+\dfrac{1}{\sqrt{c^2+1}}=2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\)
# cm: \(2\sqrt{\dfrac{c}{c+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\dfrac{3}{\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{2c\left(c^2+1\right)}+\sqrt{2c+2}\le3\sqrt{\left(c+1\right)\left(c^2+1\right)}\)
\(\Leftrightarrow8c^3+10c+2+8\sqrt{c\left(c+1\right)\left(c^2+1\right)}\le9\left(c^3+c^2+c+1\right)\)
hay \(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le c^3+9c^2-c+7\) ($)
Áp dụng BĐT AM-GM cho VT của ($):
\(8\sqrt{\left(c^2+c\right)\left(c^2+1\right)}\le4\left(2c^2+c+1\right)\) .Ta chứng minh
\(8c^2+4c+4\le c^3+9c^2-c+7\) hay \(\left(c-1\right)^2\left(c+3\right)\ge0\) (đúng)
Vậy ta có đpcm. Dấu = xảy ra khi a=b=c=1
