About an inequality in inner product space with the best approximation property I tried to prove an inequality in an inner product space $X$ which is as follows,
Let $M$ be a linear set in an inner product space $X$, and $x$$∈$$X$ such that $ρ$ = $inf$$_y$$_∈$$_M$ $\|$$x$$-$$y$$\|$. Prove that for any $y$$_1$, $y$$_2$ $∈$ $M$, $\|$$y$$_1$$-$$y$$_2$$\|$ $\leq$ $($$\|$x$-$$y$$_1$$\|$$^2$$-$$ρ$$^2$$)$$^1$$^/$$^2$ $+$$($$\|$x$-$$y$$_2$$\|$$^2$$-$$ρ$$^2$$)$$^1$$^/$$^2$.
I tried to use the property of inner product and best approximation but somewhere lost in the middle and couldn't get the desired result. How should I proceed?
 A: I’ll assume that we consider a real inner product space and linearily of the set $M$ means that if $M$ contains two distinct points $y_1$ and $y_2$ then it also contains a straight line $\ell$ going through these points. To prove the required inequality consider a triangle with vertices $x$, $y_1$, and $y_2$.  If $y_1=y_2$  then the inequality holds because its left hand side equals zero. Otherwise let $\ell$ be a straight line going through the points $y_1$ and $y_2$. Consider a height $xy_0$ of the triangle $xy_1y_2$ (that is the point $y_0$ belongs to the straight line $\ell$ and $(x-y_0,y_1-y_0)= (x-y_0,y_2-y_0)=0$).  Also $(x-y_0,x-y_0)\ge \rho^2$. Thus we have 
$$\| x-y_1\|^2-\rho^2\ge$$ $$(x-y_1, x-y_1)- (x-y_0,x-y_0)=$$ $$-2(x,y_1)+(y_1,y_1)+2(x,y_0)-(y_0,y_0)=$$
$$2(x,y_0-y_1) +(y_1,y_1) -(y_0,y_0)=$$ $$2(y_0,y_0-y_1)+(y_1,y_1) -(y_0,y_0)=$$ $$ (y_0,y_0) -2(y_0,y_1) +(y_1,y_1) =$$ $$\|y_1-y_0\|^2.$$
Similarly  we  can show that $\| x-y_2\|^2-\rho^2\ge \|y_2-y_0\|^2$. 
By triangle inequality, 
$$(\| x-y_1\|^2-\rho^2)^{1/2}+(\| x-y_2\|^2-\rho^2)^{1/2}\ge \|y_1-y_0\|+\|y_2-y_0\|\ge \|y_1-y_2\|.$$
