Question

If x ∝ 1/y, then prove that the value of (x+y) wiil be least when x=y​

Answer 1

Given,

$\longrightarrow x\propto\dfrac{1}{y}$

Let,

$\longrightarrow x=\dfrac{k}{y}$

where $k$ is a constant and does not vary with $x$ and $y.$ So,

$\longrightarrow xy=k$

By AM - GM inequality, we have,

$\longrightarrow \dfrac{x+y}{2}\geq\sqrt{xy}$

$\longrightarrow x+y\geq2\sqrt k$

Here $\sqrt k\geq0\quad\!\forall k\in\mathbb{R}.$ So $x+y$ is minimum when,

$\longrightarrow\sqrt k=0$

$\longrightarrow k=0$

$\longrightarrow xy=0$

$\Longrightarrow x=0\quad OR\quad y=0\quad\quad\dots(1)$

And so,

$\longrightarrow x+y=2\sqrt k$

$\longrightarrow x+y=0$

$\longrightarrow x=-y$

So (1) follows,

$\longrightarrow -y=0\quad OR\quad -x=0$

$\longrightarrow y=0\quad OR\quad x=0$

However, we get that $x+y$ will be the least when $x=y=0.$

Hence Proved!