# $\langle x,y \rangle (\lVert x \rVert + \lVert y \rVert) \leq \lVert x \rVert \lVert y \rVert \lVert x+y \rVert$ [duplicate]

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.

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

Proof:

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

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

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

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

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

• very beautifull Sep 20, 2020 at 15:04
• but this is only for real inner products, that is $\langle x,y \rangle \in \mathbb R$ Sep 20, 2020 at 17:56
• Indeed the OP asked us to prove the inequality in $\mathbb{R}^n$. Sep 20, 2020 at 18:02
• The property $$\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)\leq\lVert x\rVert\lVert y\rVert\lVert x+y\rVert$$ is false for complex scalar products. Indeed, if $\;x=(1+i,1+i)\in\mathbb{C}^2$, $\;y=(1-i,1-i)\in\mathbb{C}^2$ and $\;\langle x,y\rangle=x_1\overline{y_1}+x_2\overline{y_2}\in\mathbb{C}\;$ is the complex scalar product on $\mathbb{C}^2\;,\;$ it results that $\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)=4\cdot(2+2)>2\cdot2\cdot2\sqrt{2}=$ $=\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$ Sep 21, 2020 at 15:39
• The property $$\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)\le\lVert x\rVert\lVert y\rVert\lVert x+y\rVert$$ is also false even for real scalar products. Indeed, if $x=(1,0)\in\mathbb{R}^2\;$, $\;y=(-1,1)\in\mathbb{R}^2\;$ and $\langle x,y\rangle=x_1y_1+x_2y_2\in\mathbb{R}\;$ is the real scalar product on $\mathbb{R}^2\;,\;$ it results that $\;\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)=1\cdot(1+\sqrt{2})>1\cdot\sqrt{2}\cdot1=$ $=\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$ Sep 21, 2020 at 16:55

$$\newcommand{scal}[2]{\left\langle{#1};{#2}\right\rangle}\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$$\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.

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'.