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Question Number 200005 by Mingma last updated on 12/Nov/23

Answered by deleteduser1 last updated on 12/Nov/23

$${\left(\frac{a^2+b^2+c^2}{3}\right)}^{\frac{1}{2}}⩾\left(\frac{a+b+c}{3}\right)\Rightarrow a+b+c⩽3$$$$\mathrm{\Sigma }\frac{1}{2a+b}⩾\frac{9}{3(a+b+c)}⩾\frac{9}{3\times 3}=1$$$$Equalityholdswhena=b=c=1$$

Commented by Mingma last updated on 12/Nov/23

Nice. solution

Answered by mnjuly1970 last updated on 12/Nov/23

$${(a+b+c)}^2\stackrel{\mathrm{c}-\mathrm{s}}{⩽}3.(a^2+b^2+c^2)=9$$$$(a+b+c)⩽3\Rightarrow \frac{1}{a+b+c}⩾\frac{1}{3}$$$$t_2-lemma:\mathrm{I}=\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}⩾\frac{{(1+1+1)}^2}{3a+3b+3c}$$$$=\frac{9}{3(a+b+c)}⩾3.\frac{1}{3}=1\Rightarrow \mathrm{I}⩾1$$

Commented by Mingma last updated on 12/Nov/23

Nice solution

Answered by witcher3 last updated on 28/Nov/23

$$\frac{1}{2\mathrm{a}+\mathrm{b}}+\frac{1}{2\mathrm{b}+\mathrm{c}}+\frac{1}{2\mathrm{c}+\mathrm{a}}⩾\frac{9}{3(\mathrm{a}+\mathrm{b}+\mathrm{c})}$$$${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2⩾\frac{1}{3}{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2...\mathrm{just}\mathrm{cauchy}\mathrm{shwartz}$$$$\mathrm{cauchy}\mathrm{shwaftz}\mid \mathrm{a}.1+\mathrm{b}.1+\mathrm{c}.1\mid ⩽\sqrt{1^2+1^2+{\mathrm{c}}^2}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}$$$$\Rightarrow \frac{1}{(\mathrm{a}+\mathrm{b}+\mathrm{c})}⩾\frac{1}{\sqrt{3({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)}}=\frac{1}{3}$$$$\Rightarrow \frac{1}{2\mathrm{a}+\mathrm{b}}+\frac{1}{2\mathrm{b}+\mathrm{c}}+\frac{1}{2\mathrm{c}+\mathrm{a}}⩾\frac{9}{3}.\frac{1}{3}=1$$$$$$

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