14.
\(\left(ab+c\right)\left(bc+a\right)\le\dfrac{1}{4}\left(ab+bc+c+a\right)^2=\dfrac{1}{4}\left(a+c\right)^2\left(b+1\right)^2\)
Tương tự:
\(\left(ab+c\right)\left(ca+b\right)\le\dfrac{1}{4}\left(b+c\right)^2\left(a+1\right)^2\)
\(\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{4}\left(a+b\right)^2\left(c+1\right)^2\)
Nhân vế với vế và khai căn:
\(\left(ab+c\right)\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(a+1\right)\left(b+1\right)\left(c+1\right)\)
Mặt khác ta có:
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\dfrac{1}{27}\left(a+b+c+3\right)^3=8\)
\(\Rightarrow\left(ab+c\right)\left(bc+a\right)\left(ca+b\right)\le\dfrac{1}{8}.8.\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Dấu "=" xảy ra khi...
15.
\(\Leftrightarrow\sum\dfrac{2a^2}{2a^2+\left(b+c-a\right)^2}\le2\)
\(\Leftrightarrow\sum\left(\dfrac{2a^2}{2a^2+\left(b+c-a\right)^2}-1\right)\le2-3\)
\(\Leftrightarrow\sum\dfrac{\left(b+c-a\right)^2}{2a^2+\left(b+c-a\right)^2}\ge1\)
Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\\c=\dfrac{x+y}{2}\end{matrix}\right.\)
\(VT=\sum\dfrac{x^2}{2\left(\dfrac{y+z}{2}\right)^2+x^2}=\sum\dfrac{2x^2}{2x^2+\left(y+z\right)^2}\ge\sum\dfrac{2x^2}{2x^2+2\left(y^2+z^2\right)}=1\)
Dấu "=" xảy ra khi...
16.
\(\sum\dfrac{\left|a-b\right|}{\sqrt{2ab+c^2}}=\sum\dfrac{\sqrt{\left(a-b\right)^2}}{\sqrt{2ab+c^2}}=\sum\dfrac{2\left(a-b\right)^2}{2\sqrt{\left(a-b\right)^2\left(2ab+c^2\right)}}\)
\(\ge\sum\dfrac{2\left(a-b\right)^2}{\left(a-b\right)^2+2ab+c^2}=\dfrac{2\left(a-b\right)^2+2\left(b-c\right)^2+2\left(c-a\right)^2}{a^2+b^2+c^2}\)
\(=\dfrac{2\left(a^2+b^2+c^2\right)+2\left(a^2+b^2+c^2-2ab-2bc-2ca\right)}{a^2+b^2+c^2}\)
\(=\dfrac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\)
