Question

{[b² + c²] / [b + c]} + {[c² + a²] / [c + a]} + {[a² + b²] / [a + b]} is

• A. $>2(a+b+c)$
• B. $>a+b+c$
• C. $>ab+bc+ca$
• D. None of these

Answer 1

The correct option is B

$>a+b+c$

Explanation for correct options:

Step 1: Declaring the theorem:

Given: $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}$

We know, $\frac{{\left(a_1\right)}^m+{\left(a_2\right)}^m+....+{\left(a_n\right)}^m}{n}>{\left(\frac{a_1+a_2+..+a_n}{n}\right)}^m$

If, $m<0$or $m>1$

Step 2: Comparing the given equation

Here, $a_1=b_1$, $a_2=c$, $m=2$

$\Rightarrow \frac{b^2+c^2}{2}>{(\frac{b+c}{2})}^2$

$\Rightarrow \frac{b^2+c^2}{{(b+c)}^2}>\hspace{0.17em}\frac{1}{2}$

$\Rightarrow \frac{b^2+c^2}{b+c}>\frac{1}{2}(b+c)$ ..(1)

Again, $a_1=c$, $a_2=a$, $m=2$

$\Rightarrow \frac{c^2+a^2}{2}>{(\frac{c+a}{2})}^2$

$\Rightarrow \frac{c^2+a^2}{{(c+a)}^2}>\frac{1}{2}$

$\Rightarrow \frac{c^2+a^2}{c+a}>\frac{1}{2}(c+a)$ ..(2)

Now, $a_1=a$, $a_2=b$, $m=2$

$\Rightarrow \frac{a^2+b^2}{2}>{(\frac{a+b}{2})}^2$

$\Rightarrow \frac{a^2+b^2}{{(a+b)}^2}>\frac{1}{2}$

$\Rightarrow \frac{a^2+b^2}{a+b}>\frac{1}{2}(a+b)$ ..(3)

Step 3: Solving the equations:

Adding equation (1 + 2 + 3),

$\Rightarrow \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>\frac{1}{2}\left[\right(b+c)+(c+a)+(a+b\left)\right]$

$\Rightarrow \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>\frac{1}{2}(2a+2b+2c)$

$\Rightarrow \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>\frac{2}{2}(a+b+c)$

$\Rightarrow \frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>(a+b+c)$

Hence, Option (B) is correct.

Explanation for incorrect options:

Option (A): $>2(a+b+c)$

Since, $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>(a+b+c)$

Therefore $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}$ is not greater than $2(a+b+c)$.

Hence Option (A) is incorrect.

Option (C): $>ab+bc+ca$

Since, $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>(a+b+c)$

Therefore, $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}$ is not greater than $ab+bc+ca$

Hence Option (C) is incorrect.

Option (D): None of these

Since, $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}>(a+b+c)$

Therefore, $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}$ is greater than $a+b+c$. which is one of these options.

Hence, Option (D) is incorrect.

Hence, Option (B) is the correct option.