Write as $f(x)=\frac{(x-7)^2+(x+7)^2}{(x+7)^2} = 1+\frac{(x-7)^2}{(x+7)^2}$ so in $x+7>0$ there is a local min of $1$ at $x=7$ In $x+7<0$ there are no local minima and $f(x)>2$ everywhere. Hence the global minimum is 1

Let $x \in R-\{-7\}.$Prove that $$\frac{x^2 + 98}{(x + 7)^2}\geq\frac{2}{3}$$$$\frac{x^2 + 9}{(x + 7)^2}\geq\frac{9}{58}$$$$\frac{x^2 + 6}{(x + 7)^2}\geq\frac{6}{55}$$$$\frac{x^2 +x+ 1}{(x + 7)^2} \geq\frac{3}{172}$$

sqing wrote: Let $x \in R-\{-7\}.$Prove that $$\frac{x^2 + 98}{(x + 7)^2}\geq\frac{2}{3}$$$$\frac{x^2 + 9}{(x + 7)^2}\geq\frac{9}{58}$$$$\frac{x^2 + 6}{(x + 7)^2}\geq\frac{6}{55}$$ Cross multiply and use the formula for minimum value of a quadratic.

$\frac{x^2+98}{(x+7)^2}-\frac23=\frac{(x-14)^2}{3(x+7)^2}$ $\frac{x^2+9}{(x+7)^2}-\frac9{58}=\frac{(7x-9)^2}{58(x+7)^2}$ $\frac{x^2+6}{(x+7)^2}-\frac6{55}=\frac{(7x-6)^2}{55(x+7)^2}$

Sure. I'll do the first one. Cross multiplying gives $3x^2+294=2(x^2+14x+49)$, or $3x^2+294 \geq 2x^2+28x+98$, or $x^2-28x+196 = (x-14)^2 \geq 0$. Didn't even need minimum value Alternatively minimum value is attained at $x=\frac{-(-28)}{2}=14$ which gives $0$ so $x^2-28x+196 \geq 0$ as desired.