Problem-2

Question

If \(\displaystyle x,y\) are real numbers such that \(\displaystyle x^{2} +y^{2} =1\), show that \(\displaystyle -\sqrt{2} \leqslant x+y\leqslant \sqrt{2}\).

Algebraic Solution

Getting rid of the square roots, we see that we have an equivalent problem:

\[ \begin{equation*} -\sqrt{2} \leqslant x+y\leqslant \sqrt{2} \Longleftrightarrow ( x+y)^{2} \leqslant 2 \end{equation*} \]

Expanding on this:

\[ \begin{equation*} \begin{aligned} & & ( x+y)^{2} & \leqslant 2\\ \Longleftrightarrow & & x^{2} +y^{2} +2xy & \leqslant 2\\ \Longleftrightarrow & & 1+2xy & \leqslant 2\\ \Longleftrightarrow & & xy & \leqslant \frac{1}{2} \end{aligned} \end{equation*} \]

The \(\displaystyle \Longleftrightarrow\) shows that the steps are reversible. That is, if we start with \(\displaystyle xy\leqslant \frac{1}{2}\), we can work backwards and show that \(\displaystyle ( x+y)^{2} \leqslant 2\). The forward direction should be quite clear.

We have used \(\displaystyle ( x+y)^{2}\). Let us now turn to \(\displaystyle ( x-y)^{2}\) and see how this helps. To begin with, the only thing we know is that this quantity is positive:

\[ \begin{equation*} \begin{aligned} & & ( x-y)^{2} & \geqslant 0\\ \Longleftrightarrow & & x^{2} +y^{2} -2xy & \geqslant 0\\ \Longleftrightarrow & & 1-2xy & \geqslant 0\\ \Longleftrightarrow & & xy & \leqslant \frac{1}{2} \end{aligned} \end{equation*} \]

And that completes the proof of the inequality. To establish the equality, note that equality at the RHS holds when \(\displaystyle x=y=\frac{1}{\sqrt{2}}\) and at the LHS when \(\displaystyle x=y=\frac{-1}{\sqrt{2}}\).

Geometric solution

Geometrically, \(\displaystyle f( x) =x+y\) is a plane in 3D space. \(\displaystyle x^{2} +y^{2} =1\) is a cylinder with radius \(\displaystyle 1\) in space. The intersection of the plane and the cylinder results in a 2D curve in the plane embedded in space. What the inequality says is that the height of any point on this curve along the z-axis as measured from the XY plane is bounded between \(\displaystyle -\sqrt{2}\) and \(\displaystyle \sqrt{2}\).

The normal to the plane is the vector \(\displaystyle ( 1,1,-1)\). If you try to visualize the curve, the highest and lowest points will lie along the vector \(\displaystyle ( 1,1,2)\), which lies on the plane. Alternatively, the highest and lowest points lie on the line joining the origin and \((1, 1, 2)\). From this, we get \(\displaystyle x=y\), using which we have:

\[ \begin{equation*} 2x^{2} =1\Longrightarrow x=\frac{\pm 1}{\sqrt{2}} \end{equation*} \]

The minimum and maximum values of \(\displaystyle x+y\) are therefore \(\displaystyle -\sqrt{2}\) and \(\displaystyle \sqrt{2}\) respectively.