Đặt VT là K.
Ta có: \(6a^2+8ab+11b^2=\left(2a+3b\right)^2+2\left(a-b\right)^2\ge\left(2a+3b\right)^2\)
\(\Rightarrow\frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\le\frac{a^2+3ab+b^2}{2a+3b}\)
Tiếp tục ta chứng minh: \(\frac{a^2+3ab+b^2}{2a+3b}\le\frac{3a+2b}{5}\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)
Tương tự ta có: \(\frac{b^2+3bc+c^2}{\sqrt{6b^2+8bc+11c^2}}\le\frac{3b+2c}{5}\);\(\frac{c^2+3ca+a^2}{\sqrt{6c^2+8ca+11a^2}}\le\frac{3c+2a}{5}\)
Cộng từng vế của các bđt trên, ta được:
\(M\le\frac{3b+2c}{5}+\frac{3a+3b}{5}+\frac{3c+2a}{5}=a+b+c\)
Lại có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\le a^2+b^2+c^2+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
hay \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow a+b+c\le3\)
Vậy \(M\le3\)
Đẳng thức xảy ra khi a = b = c = 1
VT là M nha, mà k hay M gì cx đc, cm đc ròi
Ta có \(6a^2+8ab+11b^2=\left(2a+3b\right)^2+2\left(a-b\right)^2\ge\left(2a+2b\right)^2\)
\(\Rightarrow\frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\le\frac{a^2+3ab+b^2}{2a+3b}\)
Mặt khác \(\frac{a^2+3ab+b^2}{2a+3b}=\frac{5a^2+15ab+5b^2}{5\left(2a+3b\right)}\)
\(=\frac{\left(3a+2b\right)\left(2a+3b\right)-\left(a^2-2ab+b^2\right)}{5\left(2a+b\right)}=\frac{3a+2b}{5}-\frac{\left(a-b\right)^2}{5\left(2a+b\right)}\le\frac{3a+2b}{5}\)
Do đó \(\frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\le\frac{3a+2b}{5}\left(1\right)\)
Tương tự \(\hept{\begin{cases}\frac{b^2+3bc+c^2}{\sqrt{6b^2+8bc+11c^2}}\le\frac{3b+2c}{5}\left(2\right)\\\frac{c^2+3ca+a^2}{\sqrt{6c^2+8bc+11a^2}}\le\frac{3c+2a}{5}\left(3\right)\end{cases}}\)
Từ (1) (2)(3) ta có: \(\frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}+\frac{b^2+3bc+c^2}{\sqrt{6b^2+8bc+11c^2}}+\frac{c^2+3ca+a^2}{\sqrt{6c^2+8ca+11a^2}}\le a+b+c=3\)
