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Question Number 178697 by infinityaction last updated on 20/Oct/22

Commented by mr W last updated on 21/Oct/22

$$fora,b,c\in \mathbb{R},ithink$$$$a+b+c⩽2\sqrt{3}$$

Commented by mr W last updated on 21/Oct/22

$$ihavebeingfollowingthisquestion$$$$forsomedays.doyouhavetheright$$$$solution?$$

Commented by mr W last updated on 21/Oct/22

$$igotthesameresult,asexactvalue.$$$$aboveimistook{(a+b+c)}_{max}=2\sqrt{3}.$$$$butitisonlyalocalmaximumfrom$$$$k^2=\lambda =12.$$$$seesolutionbelow.$$

Answered by mr W last updated on 22/Oct/22

$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3=9$$$$(a+b+c)(ab+bc+ca)=12abc$$$$leta+b+c=k$$$$abc=\frac{k(ab+bc+ca)}{12}$$$${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$$$$k^2=12+2(ab+bc+ca)$$$$ab+bc+ca=\frac{k^2-12}{2}$$$$abc=\frac{k(k^2-12)}{24}$$$$a,b,carerootsoffollowingcubiceqn.$$$$z^3-{kz}^2+\frac{k^2-12}{2}z-\frac{k(k^2-12)}{24}=0$$$$sincea,b,c\in R,theeqn.musthave$$$$threerealroots.wetransformitto$$$$t^3+3pt+2q=0witht=z-\frac{k}{3}$$$$p=\frac{k^2-36}{18}$$$$q=\frac{k(11k^2-324)}{432}$$$$p^3+q^2⩽0toget3realroots$$$$32{(k^2-36)}^3+k^2{(11k^2-324)}^2⩽0$$$$let{k}^2=\lambda$$$$(\lambda -12)(17{\lambda }^2-972\lambda +13824)⩽0$$$$k_{max}^2=\lambda =\frac{486+6\sqrt{33}}{17}$$$$k_{max}=\sqrt{\frac{486+6\sqrt{33}}{17}}$$$$i.e.{(a+b+c)}_{max}=\sqrt{\frac{486+6\sqrt{33}}{17}}\approx 5.533148$$

Commented by infinityaction last updated on 22/Oct/22

$$thankssir$$

Commented by Tawa11 last updated on 23/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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