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If a, b, c,...

Question

If $a, b, c, d, e, f$ are positive real numbers such that $a + b + c + d + e + f = 3$ , then $x = (a + f) (b + e) (c + d)$ satisfies the relation-

0 < x \leq 1

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1 \leq x \leq 2

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2 \leq x \leq 3

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3 \leq x \leq 4

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Solution

The correct option is A $0 < x \leq 1$
Given that:
$A . M \geq G . M \geq H . M$ and $a, b, c, d, e, f > 0$
$a + b + c + d + e + f = 3$ and $x = (a + f) (b + e) (c + d)$
Now,
$\frac{(a + f) + (b + e) + (c + d)}{3} \geq {(a + f) (b + e) (c + d)}^{\frac{1}{3}}$
or, $\frac{3}{3} \geq x^{\frac{1}{3}}$
or, $x \leq 1$
$x > 0$ (Given that)
Therefore, $0 < x \leq 1$ satisfies the above relation.
Hence, A is the correct option.

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If a,b,c,d,e,f are positive real numbers such that a+b+c+d+e+f=3, then x=(a+f)(b+e)(c+d) satisfies the relation-

The correct option is A 0<x≤1Given that:A.M≥G.M≥H.M and a,b,c,d,e,f>0 a+b+c+d+e+f=3 and x=(a+f)(b+e)(c+d)Now,(a+f)+(b+e)+(c+d)3≥{(a+f)(b+e)(c+d)}13or, 33≥x13or, x≤1 x>0 (Given that)Therefore, 0<x≤1 satisfies the above relation.Hence, A is the correct option.

If $a, b, c, d, e, f$ are positive real numbers such that $a + b + c + d + e + f = 3$ , then $x = (a + f) (b + e) (c + d)$ satisfies the relation-