Application of Cauchy-Schwarz Inequality


The following problem, due to Professor Dorin Marghidanu, has been posted at the CutTheKnotMath faceboook page by Leo Giugiuc, along with a solution (Solution 1 below) by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

If $a,b,c\ge 1,$ prove that



Cauchy-Schwarz Inequality

The two solutions below invoke the most important and useful mathematical tool - the Cauchy-Schwarz inequality that was covered almost in passing at the old and by now dysfunctional Cut-The-Knot forum. Below I state the inequality and give two proofs (out of a known great variety.)

For all real $x_i,$ $y_i,$ $i=1,2,\ldots,n,$


The equality is only attained when the two sequences (vectors) $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ are linearly dependent, i.e., when, say, there are $u$ and $v$ such that $ux_i+vy_i=0,$ for all $i,$ $1\le i\le n.$

Proof 1


$\displaystyle\begin{align} f(t)&=\sum_{i=1}^{n}(tx_i+y_i)^2\\ &=\sum_{i=1}^{n}(t^2x_i^2+2tx_iy_i+y_i^2)\\ &=t^{2}\left(\sum_{i=1}^{n}x_i^2\right)+2t\left(\sum_{i=1}^{n}x_iy_i\right)+\sum_{i=1}^{n}y_i^2. \end{align}$

Since $f(t)\ge 0,$ for all $t\in\mathbb{R},$ the discriminant $D=\left(\sum_{i=1}^{n}x_iy_i\right)^2-\left(\sum_{i=1}^{n}x_i^2\right)\left(\sum_{i=1}^{n}y_i^2\right)$ is not positive. This is exactly the Cauchy-Schwarz inequality.

Proof 2

The Cauchy-Schwarz inequality is a direct consequence of a stronger result, Lagrange's identity:

$\displaystyle\left(\sum_{i=1}^{n}x_{i}^{2}\right)\left(\sum_{i=1}^{n}y_{i}^{2}\right)-\left(\sum_{i=1}^{n}x_iy_i\right)^{2}=\sum_{1\le i\lt j\le n}(x_iy_j-x_jy_i)^2.$

Solution 1

We denote $x=\sqrt{a^2-1}, etc.$ Obviously, $x,y,z\ge 0$ and the task becomes to prove


By the Cauchy-Schwarz inequality, $\sqrt{(x^2+1)(y^2+1)}\ge x+y,$ with equality only when $xy=1.$ Applying this term-by-term yields the required inequality. Equality holds only if $x=y=z=1,$ i.e., $a=b=c=\sqrt{2}.$

For a generalization,


where $a_{n+1}=a_1.$ For $n$ odd, the equality is only attained when all $a_i=\sqrt{2};$ for $n$ even, whenever $x_1x_2=x_2x_3=\ldots=x_nx_1=1.$

Solution 2

We'll go directly to a general case. By the Cauchy-Schwarz inequality,

$\displaystyle\begin{align} \left(\sum_{i=1}^{n}\sqrt{a_i^2-1}\right)^2 &\le n\sum_{i=1}^{n}(a_i^2-1)\\ &=n\sum_{i=1}^{n}a_i^2-n^2\\ &\le n\sum_{i=1}^{n}a_ia_{i+1}-n^2, \end{align}$

where $a_{n+1}=a_1.$ Note that the first inequality becomes equality whenever all $\sqrt{a_i^2-1}$ are equal (i.e., whenever all $a_i^2$ are equal.) The second unequality becomes equality whenever all $a_i$ are equal. Thus the required inequality will be proved if we manage to prove

$\displaystyle n\sum_{i=1}^{n}a_ia_{i+1}-n^2\le\left(\frac{1}{2}\sum_{i=1}^{n}a_ia_{i+1}\right)^2.$

Denote $t=\displaystyle\sum_{i=1}^{n}a_ia_{i+1}.$ We need to prove that

$f(t)=\left(\frac{1}{2}t\right)^2-nt+n^2\ge 0.$

This will be true for all $t\in\mathbb{R}$ provided the discriminant $D=(2n)^2-4n^2$ of the quadratic form $t^2-4nt+4n^2$ is not positive. But as a matter of fact, it is always zero, implying $f(t)\ge 0,$ with $f(2n)=0.$ It follows that

$\displaystyle n\sum_{i=1}^{n}a_ia_{i+1}-n^2\le\left(\frac{1}{2}\sum_{i=1}^{n}a_ia_{i+1}\right)^2.$

The equality is attained whenever $t=\displaystyle\sum_{i=1}^{n}a_ia_{i+1}=2n$ and all $a_i$ are equal, implying $a_i=\sqrt{2}.$


What do we learn from the above? Two solutions to the same problem, both using the Cauchy-Schwarz inequality, and, in the original problem (of three terms) producing the same results. However, the two methods lead to different generalizations for an increased number $n$ of terms. The difference is only noticeable when $n$ is even, and the second solution gives no clue that there might be a difference between the cases of odd and even number of terms. The only thing that comes to mind is that occasionally doing everything right may not necessarily yield a complete (not to use the term "right") answer. In a certain sense, the application of the Cauchy-Schwarz inequality in the first solution is more refined than its application in the second solution, but who could say that without first trying both ways?


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