Exercise 1591Calculus › Multivariable calculus › Optimisation in several variablesL1Added 3 Nov 2026

Cauchy–Schwarz via optimization

#Cauchy-Schwarz#optimization#inequality

Statement

Using Lagrange multipliers, prove that $x+y\le\sqrt{2(x^2+y^2)}$ .

Hint

Maximize $x+y$ under $x^2+y^2=1$ ; multiply by the radius.

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Lagrange. $1=2\lambda x$ , $1=2\lambda y$ , so $x=y$ . On the unit circle, $x=y=1/\sqrt2$ , giving maximum $\sqrt2$ .

Scaling. For any $(x,y)\neq(0,0)$ with $r=\sqrt{x^2+y^2}$ , $(x/r,y/r)$ is on the unit circle, so $x+y\le\sqrt2\,r$ .

Conclusion. $x+y\le\sqrt{2(x^2+y^2)}$ .