If $x^2 + y^2 = 1$, how do you prove $x+y≤\sqrt2$?
I'm really stuck (I think inequalities is involved, but I'm really bad at inequalities and rarely use them so.... ) I think C-S is involved, but I'm not sure.
This comes from a more complicated, larger problem, I've just broken it down to the bit on which I have no idea what to do.
$$(x-y)^2\ge0$$
$$⇒ x^2+y^2\ge2xy=(x+y)^2-x^2-y^2=(x+y)^2-1$$
$$⇒ 1=x^2+y^2\ge(x+y)^2-1$$
$$⇒ 2\ge(x+y)^2$$
$$⇒ x+y≤\sqrt2$$
Cauchy-Schwarz works, as you suspected. Define $$\mathbf{u}=(x,y) \\ \mathbf{v}=(1,1) $$ We have $$\mathbf{u} \cdot \mathbf{v} \leq \|\mathbf{u} \| \| \mathbf{v}\|, $$ or
$$x+y \leq \sqrt{2}.$$
$$x^2+y^2=1$$ is the unit circle, and
$$\cos(\theta)+\sin(\theta)=\sqrt2\cos\left(\theta-\dfrac\pi4\right)\le\sqrt 2.$$
Note that if $x^2+y^2=1$, then by the Arithmetic mean-Geometric mean inequality $$ |xy|=\sqrt{x^2y^2}\leqslant\frac{x^2+y^2}{2}=\frac{1}{2} $$ Now, if $x^2+y^2=1$, then $(x+y)^2=1+2xy\leqslant 1+2|xy|$, so $$ |x+y|\leqslant\sqrt{1+2|xy|}\leqslant \sqrt{1+2\cdot\frac{1}{2}}=\sqrt{2} $$ as required QED.
One could also consider a geometric perspective. The tangent line to the circle $x^2+y^2=1$ at the point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ is the line $x+y = \sqrt{2}$.
For a given point $P = (x_0,y_0)$ on the circle, if $c$ is the value of $x_0+y_0$, then the line $x+y = c$ intersects the circle in $P$. This line is parallel to the line $x+y = \sqrt{2}$, and $c$ will be less than or equal to $\sqrt{2}$.
Let us solve
$$\begin{cases}x^2+y^2=1,\\x+y=\sqrt2.\end{cases}$$
Subracting the second multiplied by $\sqrt2$,
$$\left(x-\frac1{\sqrt2}\right)^2+\left(y-\frac1{\sqrt2}\right)^2-1=1-2$$ gives the only solution $\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$.
Hence all points are on the same side of $x+y=\sqrt2$, because a disk is convex.
Assume $x+y>\sqrt2$, so that $y>\sqrt2-x$.
Then $x^2+y^2>x^2+2-2\sqrt2x+x^2=2x(x-\sqrt2)+2$.
The RHS is minimal when $x=\frac1{\sqrt2}$, and equals $1$.
Let $x+y>\sqrt{2}$
$$ x^2+y^2+2xy>2 $$
$$ 2xy>1 $$
We know that $(x-y)^2<0$ is not true.
$$ x^2-2xy+y^2 $$
$$ 1-2xy $$
Now see $2xy>1$ which will make $(x-y)^2<0$ which is false. Thus our assumption that $x+y>\sqrt{2}$ is wrong.
