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Question Number 193543 by York12 last updated on 16/Jun/23

$${solve} \\ $$

Commented by Subhi last updated on 16/Jun/23

$${cd}!!! \\ $$

Commented by York12 last updated on 16/Jun/23

$${what}\:{sir} \\ $$

Commented by Frix last updated on 16/Jun/23

$$\mathrm{Triple}\:\mathrm{faculty}\:\mathrm{of}\:{cd} \\ $$

Commented by Subhi last updated on 16/Jun/23

$${how}\:{ab}+{bc}+{cd} \\ $$$${and}\:{the}\:{given}\:{a}+{b}+{c} \\ $$$${that}\:{makes}\:{no}\:{sense} \\ $$

Commented by York12 last updated on 16/Jun/23

(√((a^2 +b^2 +c^2 )/3))≥((a+b+c)/3) ⇒((a^2 +b^2 +c^2 )/3)≥(1/9) ⇒ a^2 +b^2 +c^2 ≥(1/3) since a^2 +b^2 +c^2 ≥ab+bc+ac ⇒ the max of (ab+bc+ac)=(1/3)

$$ \\ $$$$\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}}\geqslant\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\geqslant\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${since} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ac} \\ $$$$\Rightarrow\:{the}\:{max}\:{of}\:\left({ab}+{bc}+{ac}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$

Commented by York12 last updated on 16/Jun/23

I am 8 th grade student sir and I would be happy to hear your comment if there is something wrong

$${I}\:{am}\:\mathrm{8}\:{th}\:{grade}\:{student}\:{sir}\:{and} \\ $$$${I}\:{would}\:{be}\:{happy}\:{to}\:{hear}\:{your}\:{comment} \\ $$$${if}\:{there}\:{is}\:{something}\:{wrong} \\ $$

Commented by York12 last updated on 16/Jun/23