A formal proof that maximum of square root of product of two numbers with range of 0 to 1 occurs as they both approach 1.
$\max_{x\in[0,1],y\in[0,1]}\sqrt{xy}$ as $x$ and $y$ approach 1.
Proof $\max_{x\in[0,1],y\in[0,1]}\sqrt{xy} $ when $[x=1,y=1]$
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optimization
radicals
maxima-minima
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2It is fairly clear that $\sqrt{xy} \le 1$ for all $x, y \in [0,1]$ and that when $x=y=1$, we have $ \sqrt{xy} = 1.$ Thus we have a upper bound and we found a point that meets the upper bound; this point must be the maximum. Is this not considered a formal proof? – 2017-02-08
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0@User8128, I am not sure, it is fairly intuitive to see, I wanted to make sure I am not missing any underlying formal proof. – 2017-02-08
1 Answers
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Formally, to see this, one should first make assertions on the nature of $\sqrt{xy}$.
Claim 1: (a) If $x,y,z \in [0,1]$, and $x \geq z$, then $xy \geq zy$.
(b)If $x,y,z \in [0,1]$, and $y \geq z$, then $xy \geq xz$.
Proof: $xy - zy = (x-z)y \geq 0$ as each individual term is non-negative. A similar proof holds for the second one.
Claim 2: If $a,b \in [0,\infty)$, then $a > b \iff a^2 > b^2$.
Proof: Note that $(a^2-b^2) = (a+b)(a-b)$. Since $(a+b) > 0$ is true, this equation can only hold when $a^2 -b^2$ and $a-b$ have the same sign. This is the claim.
Claim 3: $w,x,y,z \in [0,1]$, and $x \geq w, y \geq z$ implies $\sqrt{xy} \geq \sqrt{wz}$.
Proof: Note that $xy - wz = xy -yw + yw-wz = (x-w)y + w(y-z) \geq 0$. By claim $2$, we can say that $\sqrt{xy} > \sqrt{wz}$.
Theorem: $\displaystyle\sup_{x,y \in [0,1]} \sqrt{xy} = 1$, attained at $x=1,y=1$.
Proof: Let $x,y \in [0,1]$. Then, $1 \geq x, 1 \geq y$, so $\sqrt{1 \cdot 1} \geq \sqrt {xy}$. So, since the right side applies for all $x,y \in [0,1]$, it is true that $1 \geq \displaystyle\sup_{x,y \in [0,1]} \sqrt{xy}$. However, by definition of $\sup$ the reverse inequality holds, so we are done. i.e. $\displaystyle\sup_{x,y \in [0,1]} \sqrt{xy} = 1$, attained at $x=1,y=1$.