State true or false-If a1-a2-a3 - -R- then a1a2-a2a3-a3a1- 3

The correct option is A TrueGiven, a_1,a_2,a_3 ∈ +R a_1≥ 0, a_2≥ 0, a_3≥ 0 But a_1,a_2,a_3 cannot be equal to zero as they are the denominatorsTher ...

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Standard IX

Mathematics

Subtraction of Polynomials

State true or...

Question

State true or false:
If $a_{1}, a_{2}, a_{3} \in + R$ , then $\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{1}} \geq 3$

True

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False

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Solution

The correct option is A True
Given, $a_{1}, a_{2}, a_{3} \in + R ⟹ a_{1} \geq 0, a_{2} \geq 0, a_{3} \geq 0$
But $a_{1}, a_{2}, a_{3}$ cannot be equal to zero as they are the denominators
Therefore, $a_{1} \geq 1, a_{2} \geq 1, a_{3} \geq 1$

$⟹ \frac{a_{1}}{a_{2}} \geq 1, \frac{a_{2}}{a_{3}} \geq 1, \frac{a_{3}}{a_{1}} \geq 1,$

adding all the above equations we get

$⟹ \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{1}} \geq 1 + 1 + 1$

$⟹ \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{1}} \geq 3$

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Similar questions

Q. State true or false:

a_{1}, a_{2}, a_{3} \in + R

then,

(a_{1} + a_{2} + a_{3}) {(\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}^{3} \leq 27

Q. State true or false:

a_{1}, a_{2}, a_{3} \in + R

, then

3 a_{1} a_{2} a_{3} \leq a_{1}^{3} + a_{2}^{3} + a_{3}^{3}

Q. State True or False.

For positive numbers

a_{1}, a_{2}, a_{3}, . . . . . . . . ., a_{n}

;

(a_{1} + a_{2} + a_{3} + . . . . . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} . . . . . . . . . . . . + \frac{1}{a_{n}}) \geq n^{2}

Q. If

a_{1}, a_{2}, a_{3}, . . . a_{n}

are in H.P then

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

are in

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

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State true or false:If a1,a2,a3∈+R, then a1a2+a2a3+a3a1≥3

The correct option is A TrueGiven, a1,a2,a3∈+R⟹a1≥0,a2≥0,a3≥0But a1,a2,a3 cannot be equal to zero as they are the denominatorsTherefore, a1≥1,a2≥1,a3≥1⟹a1a2≥1,a2a3≥1,a3a1≥1,adding all the above equations we get⟹a1a2+a2a3+a3a1≥1+1+1⟹a1a2+a2a3+a3a1≥3

State true or false:
If $a_{1}, a_{2}, a_{3} \in + R$ , then $\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{1}} \geq 3$