If a, b, c, d are positive real numbers such | vedclass

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(A) Given $a, b, c, d > 0$ and $a + b + c + d = 2$.Let $x = a + b$ and $y = c + d$. Then $x + y = 2$.We want to find the range of $M = xy$.By the Arit

Vedclass · Accepted Answer

$0 < M \le 1$

Vedclass · Answer

(A) Given $a, b, c, d > 0$ and $a + b + c + d = 2$.Let $x = a + b$ and $y = c + d$. Then $x + y = 2$.We want to find the range of $M = xy$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that for positive real numbers $x$ and $y$,$\frac{x + y}{2} \ge \sqrt{xy}$.Substituting the values,we get $\frac{2}{2} \ge \sqrt{M}$,which implies $1 \ge \sqrt{M}$.Squaring both sides,we get $M \le 1$.Since $a, b, c, d$ are positive,$x = a + b > 0$ and $y = c + d > 0$,so $M = xy > 0$.Therefore,the relation satisfied is $0 < M \le 1$.

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation

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