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Question Number 160473 by quvonnn last updated on 30/Nov/21

	Answered by Rasheed.Sindhi last updated on 30/Nov/21
	$a+b>c ∧ b+c>a ∧ c+a>b a+b−c>0 ∧ b+c−a>0 ∧ c+a−b>0 (√(a+b−c)) ≤(√a) +(√b) −(√c) (((√(a+b−c)) )/( (√a) +(√b) −(√c)))≤1.................(i) Similarly, (((√(b+c−a)) )/( (√b) +(√c) −(√a) ))≤1.................(ii) (((√(c+a−b)) )/( (√c) +(√a) −(√b) ))≤1..................(iii) (i)+(ii)+(iii): (((√(a+b−c)) )/( (√a) +(√b) −(√c)))+(((√(b+c−a)) )/( (√b) +(√c) −(√a) ))+(((√(c+a−b)) )/( (√c) +(√a) −(√b) ))≤3 determinant ((( { ((a+b−c>0)),((b+c−a>0)),((c+a−b>0)) :} ; (_( of triangle) ^( a,b &c are sides) )),(((√(a+b)) ≤(√a) +(√b) ;a,b>0 )))$
	$a + b > c \land b + c > a \land c + a > b$ $a + b - c > 0 \land b + c - a > 0 \land c + a - b > 0$ $\sqrt{a + b - c} ⩽ \sqrt{a} + \sqrt{b} - \sqrt{c}$ $\frac{\sqrt{a + b - c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} ⩽ 1 . . . . . . . . . . . . . . . . . (i)$ $S i m i l a r l y,$ $\frac{\sqrt{b + c - a}}{\sqrt{b} + \sqrt{c} - \sqrt{a}} ⩽ 1 . . . . . . . . . . . . . . . . . (i i)$ $\frac{\sqrt{c + a - b}}{\sqrt{c} + \sqrt{a} - \sqrt{b}} ⩽ 1 . . . . . . . . . . . . . . . . . . (i i i)$ $(i) + (i i) + (i i i) :$ $\frac{\sqrt{a + b - c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} + \frac{\sqrt{b + c - a}}{\sqrt{b} + \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c + a - b}}{\sqrt{c} + \sqrt{a} - \sqrt{b}} ⩽ 3$ $\begin{array}{cc} {\begin{cases} a + b - c > 0 \\ b + c - a > 0 \\ c + a - b > 0 \end{cases}; (_{o f t r i a n g l e}^{a, b & c a r e s i d e s} \\ \sqrt{a + b} ⩽ \sqrt{a} + \sqrt{b}; a, b > 0 \end{array}$
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