Menu Close

If-a-1-b-1-c-1-8-Then-a-2-b-2-c-2-3-




Question Number 205422 by hardmath last updated on 20/Mar/24
If  (a + 1)(b + 1)(c + 1) = 8  Then:  a^2  + b^2  + c^2  ≥ 3
$$\mathrm{If} \\ $$$$\left(\mathrm{a}\:+\:\mathrm{1}\right)\left(\mathrm{b}\:+\:\mathrm{1}\right)\left(\mathrm{c}\:+\:\mathrm{1}\right)\:=\:\mathrm{8} \\ $$$$\mathrm{Then}: \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\geqslant\:\mathrm{3} \\ $$
Answered by A5T last updated on 20/Mar/24
a+1+b+1+c+1≥3(((a+1)(b+1)(c+1)))^(1/3) =6  ⇒a+b+c≥3  (√3)(√(a^2 +b^2 +c^2 ))≥a+b+c⇒a^2 +b^2 +c^2 ≥3
$${a}+\mathrm{1}+{b}+\mathrm{1}+{c}+\mathrm{1}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\left({c}+\mathrm{1}\right)}=\mathrm{6} \\ $$$$\Rightarrow{a}+{b}+{c}\geqslant\mathrm{3} \\ $$$$\sqrt{\mathrm{3}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\geqslant{a}+{b}+{c}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{3} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *