$\langle x,y \rangle (\lVert x \rVert + \lVert y \rVert) \leq \lVert x \rVert \lVert y \rVert \lVert x+y \rVert$

Question

How I can prove this inequality in $\mathbb{R}^n$?

$$\langle x,y \rangle (\lVert x \rVert + \lVert y \rVert) \leq \lVert x \rVert \lVert y \rVert \lVert x+y \rVert.$$

Intuitively it is true, because it means that

$$\cos\theta\leq\dfrac{\lVert x+y\rVert}{\lVert x\rVert+\lVert y\rVert},$$

where $\theta$ is the angle between $x$ and $y$, but I can prove this algebraically.

Answer 1

$\langle x,y\rangle(\lVert x\rVert+\lVert y\rVert)\leq\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$

Proof:

$\langle x,y\rangle(\lVert x\rVert+\lVert y\rVert)=\lVert x\rVert\langle x,y\rangle+\lVert y\rVert\langle x,y\rangle\le$

$\underset{\overbrace{\text{Cauchy–Schwarz inequality}}}{\le}\lVert x\rVert^2\lVert y\rVert+\lVert y\rVert\langle x,y\rangle=$

$\underset{\overbrace{\lVert x\rVert^2=\langle x,x\rangle}}{=}\lVert y\rVert\left(\langle x,x\rangle+\langle x,y\rangle\right)=$

$\underset{\overbrace{\text{Bilinearity of scalar product}}}{=}\lVert y\rVert\langle x,x+y\rangle\le$

$\underset{\overbrace{\text{Cauchy–Schwarz inequality}}}{\le}\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$

Answer 2

Assume $\|x\|=1$ and $y= ax+bz$ where $z\bot x$ and $\|z\|=1$.

Then the left side would be $a(1+\sqrt {a^2+b^2})$ and the right side would be$\sqrt{a^2+b^2}\sqrt{(a+1)^2+b^2}$.

We see inequality holds when $b^2=0$. Derivation shows left'$\leq$ right'.

Answer 3

$\newcommand{scal}[2]{\left\langle{#1};{#2}\right\rangle}\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$\begin{align}&\scal xy\left(\nrm x+\nrm y\right)\le \nrm x\nrm y\nrm{x+y}\Leftrightarrow\\&\scal xy<0\lor \begin{cases}\scal xy\ge0\\ \scal xy^2(\nrm x^2+\nrm y^2+2\nrm x\nrm y)\le \nrm x^2\nrm y^2\nrm {x+y}^2\end{cases}\Leftrightarrow\\ &\scal xy<0\lor \begin{cases}\scal xy\ge0\\ \scal xy^2(\nrm x^2+\nrm y^2+2\nrm x\nrm y)\le \nrm x^2\nrm y^2(\nrm x^2+\nrm y^2+2\scal xy)\end{cases}\Leftrightarrow\\ &\scal xy<0\lor \begin{cases}\scal xy\ge0\\ (\scal xy^2-\nrm x^2\nrm y^2)(\nrm x^2+\nrm y^2)+2\nrm x\nrm y\scal xy(\scal xy-\nrm x\nrm y)\le 0\end{cases}\end{align}

And by $\lvert \scal xy\rvert\le \nrm x\nrm y$ it is clear that that long LHS is $\le 0$ when $\scal xy\ge0$. Therefore the condition ends up being $\scal xy<0\lor \scal xy\ge0$, i.e. always.