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Question Number 70598 by Shamim last updated on 06/Oct/19

$$\mathrm{If}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{\Re }\mathrm{and}\frac{{\mathrm{a}}^2}{\mathrm{b}+\mathrm{c}}+\frac{{\mathrm{b}}^2}{\mathrm{c}+\mathrm{a}}+\frac{{\mathrm{c}}^2}{\mathrm{a}+\mathrm{b}}=\frac{12}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$$$$,\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{4}{3}\mathrm{then}\mathrm{find}\mathrm{a}+\mathrm{b}+\mathrm{c}=?$$

Answered by som(math1967) last updated on 06/Oct/19

$$\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{c+a}+c-(a+b+c)$$$$=\frac{12}{a+b+c}$$$$\frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b}$$$$-(a+b+c)=\frac{12}{a+b+c}$$$$(a+b+c)(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1)=\frac{12}{a+b+c}$$$$(a+b+c)(\frac{4}{3}-1)=\frac{12}{a+b+c}★$$$$\frac{{(a+b+c)}^2}{3}=12$$$${(a+b+c)}^2=36\Rightarrow (a+b+c)=\pm 6$$$$★\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{4}{3}$$

Commented by Shamim last updated on 06/Oct/19

$$\mathrm{tnks}$$

Commented by Rasheed.Sindhi last updated on 06/Oct/19

$$\vee \sqcap \mid ⊏\in !$$

Commented by som(math1967) last updated on 06/Oct/19

$$Niceuseofmathematicalsymbols$$

Commented by peter frank last updated on 06/Oct/19

$$good$$

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