Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that $$\left(\frac12(x+y)\right)^2 \ge 0 $$ but stumped on where to go from here...
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1$\begingroup$ Compute the RHS minus the LHS, this is a perfect square. $\endgroup$ – Did Feb 5 '14 at 21:39
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$\begingroup$ This is basically the simplest case of the inequality between arithmetic and quadratic mean (for two variables). $\endgroup$ – Martin Sleziak May 16 '14 at 21:41
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Proof by picture (angle H is 90°).
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Expand both sides and we have:
$$\frac{x^2 + y^2}{2} \ge \frac{x^2 + 2xy + y^2}{4}$$ $$\iff x^2 + y^2 \ge {2xy}$$ $$\iff (x-y)^2 \ge 0$$
Which is well-know fact.
Also there are lot of other ways to prove it, here's "fancy" one using Cauchy-Scwarz inequality:
$$(1 + 1)(x^2 + y^2) \ge (x + y)^2$$
Multiply both sides by $\frac 14$:
$$\frac 12 (x^2 + y^2) \ge \left(\frac 12 (x+y)\right)^2$$
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$\begingroup$ What norm and wich vectors are you applying C-S on? I'm somewhat confused - sorry if it's trivial^^ // Okay see it now. It's $\mathbb R^2$ and the vectors $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}x\\y\end{pmatrix}$ $\endgroup$ – AlexR Feb 5 '14 at 21:57
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$\begingroup$ @AlexR To be fair I'm not familiar too much with vector use of Cauchy-Schwarz, but I think that those are the vectors used. As a contest participant I am more familiar with the use of real values: $$\left(\sum_{n=0}^k x_k^2\right) \cdot \left(\sum_{n=0}^k y_k^2\right) \ge \left(\sum_{n=0}^k x_ky_k\right)^2$$ In this case $x_1 = x_2 = 1$ and $y_1 = x^2$ and $y_2 = y^2$ $\endgroup$ – Stefan4024 Feb 5 '14 at 22:27
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$\begingroup$ You mean $y_1=x, y_2 = y$ ;) The vector form is more general (because it works on all inner product spaces, not only $\mathbb R^n$) $\endgroup$ – AlexR Feb 5 '14 at 22:29
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$\begingroup$ @AlexR Yes, you are right $y1=x, y_2 = y$. I might consider getting more familiar with vector form, because as you've mentioned it has much more wider use. $\endgroup$ – Stefan4024 Feb 5 '14 at 22:40
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$\frac12 (x^2+y^2)-(\frac12 (x+y))^2$
$=\frac12 (x^2+y^2)-\frac14 (x^2+y^2+2xy)$
$=\frac14 (2x^2+2y^2-x^2-y^2-2xy)$
$=\frac14 (x^2+y^2-2xy)$
$=\frac14 (x-y)^2\ge 0$
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This substitution is a nice way to do this:
$$x=r\sin\theta , y=r\cos\theta$$ So, our identity: $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$ $$\frac{1}{4} (x^2+y^2+2xy) \leq \frac12(x^2 + y^2)$$
Applying our substitutions:
$$\frac{1}{4} (r^2+2\cdot r\sin\theta\cdot r\cos\theta) \leq \frac12(r^2)$$
$$\frac{r^2}{2}\cdot \frac{1+\sin2\theta}{2} \leq \frac{r^2}{2}$$
Now, if you note that the maximum value of $\sin 2\theta$ is $1$, then it becomes obvious.
Added: It could have also been done it this way: $$\left(\frac{1}{2}(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$ $$\frac{1}{2}\cdot\frac{1}{2}(r\cos\theta+r\sin\theta)^2 \leq \frac{r^2}{2}$$ $$\frac{r^2}{2}\cdot\frac{1}{2}(\cos\theta+\sin\theta)^2 \leq \frac{r^2}{2}$$
Now, note that $-\sqrt{2} \leq \cos\theta+\sin\theta = \sqrt{2}\cos(x+45^{\circ}) \leq \sqrt{2}$. Thus the maximum value of $(\cos\theta+\sin\theta)^2$ is $2$, and the result follow.
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$$(x-y)^2\geq0$$ $$2xy\leq x^2+y^2$$ $$x^2+y^2+2xy\leq 2x^2+2y^2$$ $$(x+y)^2\leq 2x^2+2y^2$$
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let investigate $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}=\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}\leq 0.$$ So $$\frac{(x+y)^2}{4}\leq\frac{x^2+y^2}{2}.$$
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Notice that
$$(\frac{1}{2}(x+y))^2=\frac{1}{4}(x^2+2xy+y^2)\leq \frac{1}{4}(x^2+[x^2+y^2]+y^2)$$ (why?)
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