JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

a, b

and

c

are the sides of a triangle, then

\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}

View answer and explanation

Correct option: B

\geq 3

The correct answer is B (

\geq 3

) because it is proven by substituting

x=b+c-a, y=c+a-b, z=a+b-c

and applying the AM-GM inequality.

More Questions from QUADRATIC EQUATIONS

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\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

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\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

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{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)