Question Number 59627 by naka3546 last updated on 12/May/19
Answered by tanmay last updated on 12/May/19
![((3−1)/3)[(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))] =(((b+c)/a^2 )+((c+a)/b^2 )+((a+b)/c^2 )+(1/a)+(1/b)+(1/c))−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) =((b+c)/a^2 )+((c+a)/b^2 )+((a+b)/c^2 )+(1/a)+(1/b)+(1/c)−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) let left hand side expression=p right hand side exoression=q we have to prove p−q=+ve ot q−p=−ve now q=p+(1/a)+(1/b)+(1/c)−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) q−p (1/a)+(1/b)+(1/c)≥3((1/(abc)))^(1/3) ((a+b+c)/3)≥(abc)^(1/3) (1/a^2 )+(1/b^2 )+(1/c^2 )≥3((1/(a^2 b^2 c^2 )))^(1/3) (a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥3(abc)^(1/3) ×3((1/(a^2 b^2 c^2 )))^(1/3) so (1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥3(abc)^(1/3) ×(1/((abc)^(2/3) )) (1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥(3/((abc)^(1/3) )) now 3((1/(abc)))^(1/3) −(3/((abc)^(1/3) )) =0 correction done so p=q ... ....](https://www.tinkutara.com/question/Q59634.png)
$$\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}}\left[\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)\right] \\ $$$$=\left(\frac{{b}+{c}}{{a}^{\mathrm{2}} }+\frac{{c}+{a}}{{b}^{\mathrm{2}} }+\frac{{a}+{b}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right) \\ $$$$=\frac{{b}+{c}}{{a}^{\mathrm{2}} }+\frac{{c}+{a}}{{b}^{\mathrm{2}} }+\frac{{a}+{b}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right) \\ $$$${let}\:{left}\:{hand}\:{side}\:{expression}={p} \\ $$$${right}\:{hand}\:{side}\:{exoression}={q} \\ $$$${we}\:{have}\:{to}\:{prove}\:{p}−{q}=+{ve} \\ $$$${ot}\:{q}−{p}=−{ve} \\ $$$${now} \\ $$$${q}={p}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right) \\ $$$${q}−{p} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\geqslant\mathrm{3}\left(\frac{\mathrm{1}}{{abc}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\geqslant\mathrm{3}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)\geqslant\mathrm{3}\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ×\mathrm{3}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${so} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)\geqslant\mathrm{3}\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ×\frac{\mathrm{1}}{\left({abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)\geqslant\frac{\mathrm{3}}{\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$${now} \\ $$$$\mathrm{3}\left(\frac{\mathrm{1}}{{abc}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\frac{\mathrm{3}}{\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$=\mathrm{0}\:\:\:\:\:{correction}\:{done} \\ $$$${so}\:{p}={q} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$… \\ $$$$…. \\ $$$$ \\ $$$$ \\ $$