Let $x,y\in R$.show that $$\color{crimson}{f(x,y)=\sqrt{x^2-2x+16}+\sqrt{y^2-14y+64} + \sqrt{x^2-16x+y^2-14y+\frac{7}{4}xy+64} \ge 11}$$
Everything I tried has failed so far.use Computer found this inequality $\color{blue}=$ iff only if $\color{blue}{x=2,y=6}$
Here is one thing I tried, but obviously didn't work. $$f(x,y)=\sqrt{(x-1)^2+15}+\sqrt{(y-7)^2+15}+\sqrt{(x-8)^2+(y-7)^2+\dfrac{7}{4}xy-49}$$
Thanks in advance
For convenience, we make the translation $x=2+a$ and $y=6+b$, so that the equality case is $a=b=0$. Then the expression to bound is: $$\sqrt{(a+1)^2+15}+\sqrt{(b-1)^2+15}+\sqrt{\frac{7}{8}(a+b)^2+\frac{1}{8}(a-6)^2+\frac{1}{8}(b+6)^2} $$ Now recall the following form of Cauchy-Schwarz for $n$ nonnegative variables $x_1, \cdots, x_n$: $$\sqrt{n(x_1+\cdots+x_n)}=\sqrt{(1+\cdots+1)(x_1+\cdots+x_n)}\ge \sqrt{x_1}+\cdots+\sqrt{x_n}$$ with equality iff $x_1=\cdots=x_n$. We use this three times, keeping in mind the equality case $a=b=0$: $$\sqrt{(a+1)^2+15}=\frac{1}{4}\sqrt{16((a+1)^2+15)}\ge \frac{1}{4}(|a+1|+15)$$ $$\sqrt{(b-1)^2+15}=\frac{1}{4}\sqrt{16((b-1)^2+15)}\ge \frac{1}{4}(|b-1|+15)$$ $$\sqrt{\frac{7}{8}(a+b)^2+\frac{1}{8}(a-6)^2+\frac{1}{8}(b+6)^2}\ge \frac{1}{4}\sqrt{2((a-6)^2+(b+6)^2)}\ge \frac{1}{4}(|a-6|+|b+6|)$$ Now since $|a-6|+|a+1|\ge 7$ and $|b+6|+|b-1|\ge 7$ by the triangle inequality, the expression must be at least $\frac{15}{2}+\frac{7}{2}=11$, as required.
