Question

Let a,b,c be non-zero real numbers such that $a+b+c=0$; let $q=a^2+b^2+c^2$ and $r=a^4+b^4+c^4$ Then

• A. $q^2<2ralways$
• B. $q^2=2ralways$
• C. $q^2>2ralways$
• D. $q^2-2rcantakebothpositiveandnegativevalue$

Answer 1

The correct option is A $q^2<2ralways$

Given $(a+b+c)=0$

$\&$ $q=a^2+b^2+c^2$ $\&$ $r=a^4+b^4+c^4$

We know,

$\Rightarrow {(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

Therefore $0^2=q+2(ab+bc+ca)$

$\Rightarrow ab+bc+ca=\frac{-q}{2}⟶\left(1\right)$

Now, ${(a^2+b^2+c^2)}^2=a^4+b^4+c^4+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$ (all squared )

$\Rightarrow q^2=r+2$ ( some positive number )

$\Rightarrow {(ab+bc+ca)}^2=a^2{b}^2+b^2{c}^2+c^2{a}^2+2(a{b}^{2}c+a^{2}bc+ab{c}^2)$

$\Rightarrow \frac{q^2}{4}-2=a^2{b}^2+b^2{c}^2+c^2{a}^2$

$\therefore a^2{b}^2+b^2{c}^2+c^2{a}^2<\frac{q^2}{4}$

$\therefore q^2<r+2\left(\frac{q^2}{4}\right)$

$\Rightarrow q^2<r+\frac{q^2}{2}$

$\Rightarrow \frac{q^2}{2}<r$

$\Rightarrow q^2<2r$

Hence, the answer is $q^2<2r.$