Question-200005

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

Answered by AST last updated on 12/Nov/23

(((a^2 +b^2 +c^2 )/3))^(1/2) ≥(((a+b+c)/3))⇒a+b+c≤3 Σ(1/(2a+b))≥(9/(3(a+b+c)))≥(9/(3×3))=1 Equality holds when a=b=c=1

$$\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\Rightarrow{a}+{b}+{c}\leqslant\mathrm{3} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{a}+{b}}\geqslant\frac{\mathrm{9}}{\mathrm{3}\left({a}+{b}+{c}\right)}\geqslant\frac{\mathrm{9}}{\mathrm{3}×\mathrm{3}}=\mathrm{1} \\ $$$${Equality}\:{holds}\:{when}\:{a}={b}={c}=\mathrm{1} \\ $$

Commented by Mingma last updated on 12/Nov/23

Nice. solution

Answered by mnjuly1970 last updated on 12/Nov/23

( a+b+c)^2 ≤^(c−s) 3 .(a^2 +b^2 +c^2 )=9 (a+b+c) ≤3 ⇒(1/(a+b+c)) ≥(1/3) t_2 −lemma: I= (1/(2a+b)) +(1/(2b+c)) +(1/(2c+a)) ≥ (((1+1+1)^2 )/(3a+3b+3c)) = (9/(3(a+b+c)))≥ 3.(1/3)=1⇒ I≥1

$$\:\:\:\:\left(\:{a}+{b}+{c}\right)^{\mathrm{2}} \overset{\mathrm{c}−\mathrm{s}} {\leqslant}\mathrm{3}\:.\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\left({a}+{b}+{c}\right)\:\leqslant\mathrm{3}\:\Rightarrow\frac{\mathrm{1}}{{a}+{b}+{c}}\:\geqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:{t}_{\mathrm{2}} \:−{lemma}:\:\mathrm{I}=\:\frac{\mathrm{1}}{\mathrm{2}{a}+{b}}\:+\frac{\mathrm{1}}{\mathrm{2}{b}+{c}}\:+\frac{\mathrm{1}}{\mathrm{2}{c}+{a}}\:\geqslant\:\frac{\left(\mathrm{1}+\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}{a}+\mathrm{3}{b}+\mathrm{3}{c}} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{9}}{\mathrm{3}\left({a}+{b}+{c}\right)}\geqslant\:\mathrm{3}.\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1}\Rightarrow\:\mathrm{I}\geqslant\mathrm{1} \\ $$

Commented by Mingma last updated on 12/Nov/23

Nice solution

Answered by witcher3 last updated on 28/Nov/23

(1/(2a+b))+(1/(2b+c))+(1/(2c+a))≥(9/(3(a+b+c))) a^2 +b^2 +c^2 ≥(1/3)(a+b+c)^2 ...just cauchy shwartz cauchy shwaftz ∣ a.1+b.1+c.1∣≤(√(1^2 +1^2 +c^2 ))(√(a^2 +b^2 +c^2 )) ⇒(1/((a+b+c)))≥(1/( (√(3(a^2 +b^2 +c^2 )))))=(1/3) ⇒(1/(2a+b))+(1/(2b+c))+(1/(2c+a))≥(9/3).(1/3)=1

$$\frac{\mathrm{1}}{\mathrm{2a}+\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{2b}+\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{2c}+\mathrm{a}}\geqslant\frac{\mathrm{9}}{\mathrm{3}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} …\mathrm{just}\:\mathrm{cauchy}\:\mathrm{shwartz} \\ $$$$\mathrm{cauchy}\:\mathrm{shwaftz}\:\mid\:\mathrm{a}.\mathrm{1}+\mathrm{b}.\mathrm{1}+\mathrm{c}.\mathrm{1}\mid\leqslant\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}\geqslant\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2a}+\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{2b}+\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{2c}+\mathrm{a}}\geqslant\frac{\mathrm{9}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1} \\ $$$$ \\ $$

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