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Question Number 200302 by Calculusboy last updated on 16/Nov/23

Answered by Rasheed.Sindhi last updated on 18/Nov/23

(√(a+bx)) +(√(b+cx)) +(√(c+ax)) =(√(b−ax)) +(√(c−bx)) +(√(a−cx)) a+bx≥0 ∧ b+cx≥0 ∧ c+ax≥0 (a+b+c)+(a+b+c)x≥0⇒x≥−((a+b+c)/(a+b+c))=−1 determinant (((x≥−1))) b−ax≥0 ∧ c−bx≥0 ∧ a−cx≥0 (a+b+c)−(a+b+c)x≥0⇒(a+b+c)x≤a+b+c determinant (((x≤1))) determinant (((−1≤x≤1))) .....

$$\sqrt{{a}+{bx}}\:+\sqrt{{b}+{cx}}\:+\sqrt{{c}+{ax}}\:\:=\sqrt{{b}−{ax}}\:+\sqrt{{c}−{bx}}\:+\sqrt{{a}−{cx}}\: \\ $$$${a}+{bx}\geqslant\mathrm{0}\:\wedge\:{b}+{cx}\geqslant\mathrm{0}\:\wedge\:{c}+{ax}\geqslant\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)+\left({a}+{b}+{c}\right){x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\frac{{a}+{b}+{c}}{{a}+{b}+{c}}=−\mathrm{1} \\ $$$$\begin{array}{|c|}{{x}\geqslant−\mathrm{1}}\\\hline\end{array} \\ $$$${b}−{ax}\geqslant\mathrm{0}\:\wedge\:{c}−{bx}\geqslant\mathrm{0}\:\wedge\:{a}−{cx}\geqslant\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)−\left({a}+{b}+{c}\right){x}\geqslant\mathrm{0}\Rightarrow\left({a}+{b}+{c}\right){x}\leqslant{a}+{b}+{c} \\ $$$$\begin{array}{|c|}{{x}\leqslant\mathrm{1}}\\\hline\end{array} \\ $$$$\begin{array}{|c|}{−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}}\\\hline\end{array} \\ $$$$..... \\ $$

Commented by Calculusboy last updated on 20/Nov/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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