If a, b, c are positive integers,then (a + b) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since the Arithmetic Mean $(AM)$ is greater than the Geometric Mean $(GM)$ for positive real numbers,we have:$\frac{a + b}{2} > \sqrt{ab}$$\frac{b

Vedclass · Accepted Answer

$> 8abc$

Vedclass · Answer

(B) Since the Arithmetic Mean $(AM)$ is greater than the Geometric Mean $(GM)$ for positive real numbers,we have:$\frac{a + b}{2} > \sqrt{ab}$$\frac{b + c}{2} > \sqrt{bc}$$\frac{c + a}{2} > \sqrt{ca}$Multiplying these three inequalities together,we get:$\left(\frac{a + b}{2}\right) \left(\frac{b + c}{2}\right) \left(\frac{c + a}{2}\right) > \sqrt{ab} \cdot \sqrt{bc} \cdot \sqrt{ca}$$\frac{(a + b)(b + c)(c + a)}{8} > \sqrt{a^2 b^2 c^2}$$\frac{(a + b)(b + c)(c + a)}{8} > abc$Therefore,$(a + b)(b + c)(c + a) > 8abc$.

If $a, b, c$ are positive integers,then $(a + b)(b + c)(c + a)$ is

Similar Questions

Let $a_1, a_2, ..., a_{10}$ be a $G.P.$ If $\frac{a_3}{a_1} = 25$,then $\frac{a_9}{a_5}$ is equal to:

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

The number of terms in the series $101 + 99 + 97 + ..... + 47$ is

If the sum of the first $n$ terms of two arithmetic progressions $3, 7, 11, 15, \ldots$ and $30, 33, 36, 39, \ldots$ are equal,then find the value of $n$.

Three numbers are selected from the set ${3^1, 3^2, 3^3, \dots, 3^{20}}$. Find the number of ways these selected numbers can form an increasing Geometric Progression $(G.P.)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module