Maximum value of - 1-b-c-a a is - where a- b- c are integral sides of a triangle- - - 2 is-are divisible by

The correct options are A 1 D 3 #160;∵ #160; a, b, c are sides of a triangleSo a gt; 0, b gt; 0, c gt; 0 and a + b gt; ...

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Standard XII

Mathematics

AM,GM,HM Inequality

Maximum value...

Question

Maximum value of $\prod {(1 + \frac{b - c}{a})}^{a}$ is $λ$ where $a, b, c$ are integral sides of a triangle, $λ + 2$ is/are divisible by

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Solution

The correct options are
A $1$
D $3$
$∵ a, b, c$ are sides of a triangle
So $a > 0, b > 0, c > 0$ and $a + b > c, b + c > a, c + a > b$
Using AM $\geq$ GM for weighted means we get
$\frac{a (1 + \frac{b - c}{a}) + b (1 + \frac{c - a}{b}) + c (1 + \frac{a - b}{c})}{a + b + c} \geq {({(1 + \frac{b - c}{a})}^{a} {(1 + \frac{c - a}{b})}^{b} {(1 + \frac{a - b}{c})}^{c})}^{1 / a + b + c}$

$∴ \prod {(1 + \frac{b - c}{a})}^{a} \leq {(\frac{a + b + c}{a + b + c})}^{a + b + c} \leq 1$

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Maximum value of ∏(1+b−ca)a is λ where a,b,c are integral sides of a triangle, λ+2 is/are divisible by

The correct options are A 1 D 3∵a,b,c are sides of a triangleSo a>0,b>0,c>0 and a+b>c,b+c>a,c+a>bUsing AM ≥ GM for weighted means we geta(1+b−ca)+b(1+c−ab)+c(1+a−bc)a+b+c≥((1+b−ca)a(1+c−ab)b(1+a−bc)c)1/a+b+c∴∏(1+b−ca)a≤(a+b+ca+b+c)a+b+c≤1

Maximum value of $\prod {(1 + \frac{b - c}{a})}^{a}$ is $λ$ where $a, b, c$ are integral sides of a triangle, $λ + 2$ is/are divisible by