Question

If $a,bandc$ are$+ve$, then $a^2+b^2+c^2$is

• A. $>a+b+c$
• B. $>ab+bc+ca$
• C. $<a+b+c$
• D. $<ab+bc+ca$

Answer 1

The correct option is B

$>ab+bc+ca$

Explanation for the correct option:

Step 1: We will apply A.M and G.M inequality on$a^2\&b^2$.

We have given $a,bandc$are $+ve$numbers.

We have to find the inequality of $a^2+b^2+c^2$

we know that $A.M\ge G.M$

For $a^2$ and $b^2$

$A.M\ge G.M$

$\begin{array}{rcl}\frac{a^2+b^2}{2}& \ge & \sqrt{a^2{b}^2}\\ a^2+b^2& \ge & 2ab........\left(1\right)\end{array}$

Step 2: Similarly, apply A.M and G.M inequality on$b^2\&c^2$ and$c^2\&a^2$.

for$b^{2}and{c}^2$,

$b^2+c^2\ge 2bc.........\left(2\right)$

for$c^{2}and{a}^2$,

$c^2+a^2\ge 2ca..........\left(3\right)$

Step 3: Adding $\left(1\right),\left(2\right)and\left(3\right),$

we get,

$\begin{array}{rcl}& \Rightarrow & 2(a^2+b^2+c^2)\ge 2(ab+bc+ca)\\ & \Rightarrow & (a^2+b^2+c^2)\ge (ab+bc+ca)\end{array}$

Hence, the option $\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathbf{(}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}^{\mathbf{2}}\mathbf{)}\mathbf{>}\mathbf{(}\mathit{a}\mathit{b}\mathbf{+}\mathit{b}\mathit{c}\mathbf{+}\mathit{c}\mathit{a}\mathbf{)}$is correct.