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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.

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3 Answers 3

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

Proof:

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

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

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

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

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

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  • $\begingroup$ very beautifull $\endgroup$
    – Physor
    Sep 20, 2020 at 15:04
  • $\begingroup$ but this is only for real inner products, that is $\langle x,y \rangle \in \mathbb R$ $\endgroup$
    – Physor
    Sep 20, 2020 at 17:56
  • $\begingroup$ Indeed the OP asked us to prove the inequality in $\mathbb{R}^n$. $\endgroup$
    – Angelo
    Sep 20, 2020 at 18:02
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    $\begingroup$ 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 $\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)=4\cdot(2+2)>2\cdot2\cdot2\sqrt{2}=$ $=\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$ $\endgroup$
    – Angelo
    Sep 21, 2020 at 15:39
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    $\begingroup$ 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 $\;\left|\langle x,y\rangle\right|(\lVert x\rVert+\lVert y\rVert)=1\cdot(1+\sqrt{2})>1\cdot\sqrt{2}\cdot1=$ $=\lVert x\rVert\lVert y\rVert\lVert x+y\rVert\;.$ $\endgroup$
    – Angelo
    Sep 21, 2020 at 16:55
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$\newcommand{scal}[2]{\left\langle{#1};{#2}\right\rangle}\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$\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.

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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'.

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