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Question-160473




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 )))
$$ \\ $$$${a}+{b}>{c}\:\wedge\:{b}+{c}>{a}\:\wedge\:{c}+{a}>{b} \\ $$$${a}+{b}−{c}>\mathrm{0}\:\wedge\:{b}+{c}−{a}>\mathrm{0}\:\wedge\:{c}+{a}−{b}>\mathrm{0} \\ $$$$\sqrt{{a}+{b}−{c}}\:\leqslant\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}} \\ $$$$\frac{\sqrt{{a}+{b}−{c}}\:}{\:\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}}\leqslant\mathrm{1}……………..\left({i}\right) \\ $$$${Similarly}, \\ $$$$\frac{\sqrt{{b}+{c}−{a}}\:}{\:\sqrt{{b}}\:+\sqrt{{c}}\:−\sqrt{{a}}\:}\leqslant\mathrm{1}……………..\left({ii}\right) \\ $$$$\frac{\sqrt{{c}+{a}−{b}}\:}{\:\sqrt{{c}}\:+\sqrt{{a}}\:−\sqrt{{b}}\:}\leqslant\mathrm{1}………………\left({iii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\frac{\sqrt{{a}+{b}−{c}}\:}{\:\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}}+\frac{\sqrt{{b}+{c}−{a}}\:}{\:\sqrt{{b}}\:+\sqrt{{c}}\:−\sqrt{{a}}\:}+\frac{\sqrt{{c}+{a}−{b}}\:}{\:\sqrt{{c}}\:+\sqrt{{a}}\:−\sqrt{{b}}\:}\leqslant\mathrm{3} \\ $$$$ \\ $$$$\begin{array}{|c|c|}{\begin{cases}{{a}+{b}−{c}>\mathrm{0}}\\{{b}+{c}−{a}>\mathrm{0}}\\{{c}+{a}−{b}>\mathrm{0}}\end{cases}\:\:\:;\:\left(_{\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{triangle}} ^{\:{a},{b}\:\&{c}\:{are}\:{sides}} \right.}\\{\sqrt{{a}+{b}}\:\leqslant\sqrt{{a}}\:+\sqrt{{b}}\:\:\:\:;{a},{b}>\mathrm{0}\:}\\\hline\end{array} \\ $$

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