Áp dụng BĐT $AM-GM$ ta có:
$a^2+1\ge2a$
$b^2+1\ge2b$
$c^2+1\ge2c$
Do đó ta có: $K=\frac{a}{\sqrt{a^2+1+6}}+\frac{b}{\sqrt{b^2+1+6}}+\frac{c}{\sqrt{c^2+1+6}}\le \frac{a}{\sqrt{2a^2+2(ab+bc+ca)}}+\frac{b}{\sqrt{2b^2+2(ab+bc+ca)}}+\frac{c}{\sqrt{2c^2+2(ab+bc+ca)}}$
$\Leftrightarrow K \le \frac{a}{\sqrt{2(a+b)(a+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}$
Tiếp tục áp dụng BĐT $AM-GM$ ta có:
$\frac{1}{2\sqrt{2}}\left ( \frac{a}{a+b}+\frac{a}{a+c} \right )\ge \frac{a}{\sqrt{2(a+b)(a+c)}}$
$\frac{1}{2\sqrt{2}}\left ( \frac{b}{a+b}+\frac{b}{b+c} \right )\ge \frac{b}{\sqrt{2(a+b)(b+c)}}$
$\frac{1}{2\sqrt{2}}\left ( \frac{c}{a+c}+\frac{c}{b+c} \right )\ge \frac{c}{\sqrt{2(a+c)(b+c)}}$
Cộng $3$ bất đẳng thức trên vế theo vế ta có:
$\frac{a}{\sqrt{2(a+b)(b+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}\le\frac{1}{2\sqrt{2}}\left ( \frac{a+b}{a+b}+\frac{c+b}{b+c}+\frac{a+c}{a+c} \right )=\frac{3}{2\sqrt{2}}$
$\Rightarrow$ $K \le \frac{a}{\sqrt{2(a+b)(a+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}\le\frac{3}{2\sqrt{2}}$
Vậy $GTLN=\frac{3}{2\sqrt{2}}$
Dấu$"="$ xảy ra khi và chỉ khi: $a=b=c=1$
Áp dụng BĐT $AM-GM$ ta có:
$a^4+1\ge2a^2$
$b^4+1\ge2b^2$
$c^4+1\ge2c^2$
Do đó ta có: $K=\frac{a}{\sqrt{a^4+1+6}}+\frac{b}{\sqrt{b^4+1+6}}+\frac{c}{\sqrt{c^4+1+6}}\le \frac{a}{\sqrt{2a^2+2(ab+bc+ca)}}+\frac{b}{\sqrt{2b^2+2(ab+bc+ca)}}+\frac{c}{\sqrt{2c^2+2(ab+bc+ca)}}$
$\Leftrightarrow K \le \frac{a}{\sqrt{2(a+b)(a+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}$
Tiếp tục áp dụng BĐT $AM-GM$ ta có:
$\frac{1}{2\sqrt{2}}\left ( \frac{a}{a+b}+\frac{a}{a+c} \right )\ge \frac{a}{\sqrt{2(a+b)(a+c)}}$
$\frac{1}{2\sqrt{2}}\left ( \frac{b}{a+b}+\frac{b}{b+c} \right )\ge \frac{b}{\sqrt{2(a+b)(b+c)}}$
$\frac{1}{2\sqrt{2}}\left ( \frac{c}{a+c}+\frac{c}{b+c} \right )\ge \frac{c}{\sqrt{2(a+c)(b+c)}}$
Cộng $3$ bất đẳng thức trên vế theo vế ta có:
$\frac{a}{\sqrt{2(a+b)(b+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}\le\frac{1}{2\sqrt{2}}\left ( \frac{a+b}{a+b}+\frac{c+b}{b+c}+\frac{a+c}{a+c} \right )=\frac{3}{2\sqrt{2}}$
$\Rightarrow$ $K \le \frac{a}{\sqrt{2(a+b)(a+c)}}+\frac{b}{\sqrt{2(b+c)(b+a)}}+ \frac{c}{\sqrt{2(c+a)(c+b)}}\le\frac{3}{2\sqrt{2}}$
Vậy $GTLN=\frac{3}{2\sqrt{2}}$
Dấu$"="$ xảy ra khi và chỉ khi: $a=b=c=1$
P/s: ở dưới mình ghi lộn $a^2$ ạ
