Prove $\sqrt{\sum^n_{k=1}(a_k+b_k)^2} \leq \sqrt{\sum^n_{k=1}(a_k)^2} + \sqrt{\sum^n_{k=1}(b_k)^2}$.

Question

I wanted to use the induction method to solve this. If n=1 then $\sqrt{(a_1+b_1)^2}= \vert a_1+b_1 \vert \leq \vert a_1 \vert + \vert b_1 \vert =\sqrt{a_1^2} + \sqrt{b_1^2}$.

If n=2, then we have to show $\sqrt{\sum^2_{k=1}(a_k+b_k)^2} \leq \sqrt{\sum^2_{k=1}(a_k)^2} + \sqrt{\sum^2_{k=1}(b_k)^2}$ . . . (1).

i.e. squaring both sides, $ (a_1+b_1)^2+(a_2+b_2)^2 \leq a_1^2+a_2^2 +b_1^2+ b_2^2+ 2 \sqrt{a_1^2b_1^2+a_2^2b_2^2+ a_2^2b_1^2+ a_1^2b_2^2}$

i.e. $ a_1^2+b_1^2+a_2^2 + b_2^2 + 2 a_1 b_1 +2 a_2 b_2 \leq a_1^2+a_2^2 +b_1^2+ b_2^2+ 2 \sqrt{a_1^2b_1^2+a_2^2b_2^2+ a_2^2b_1^2+ a_1^2b_2^2} $

i.e. $ a_1 b_1 + a_2 b_2 \leq \sqrt{a_1^2b_1^2+a_2^2b_2^2+ a_2^2b_1^2+ a_1^2b_2^2} $

i.e. $ a_1^2 b_1^2 + a_2^2 b_2^2+ 2a_1a_2b_1b_2 \leq a_1^2b_1^2+a_2^2b_2^2+ a_2^2b_1^2+ a_1^2b_2^2 $

i.e. $ 2a_1a_2b_1b_2 \leq a_2^2b_1^2+ a_1^2b_2^2 $

i.e. $(a_1b_2-a_2b_1)^2 \geq 0$, Which follows equation (1) is true.

Now let us consider given inequality is true for k=n. I like to show $\sqrt{\sum^{n+1}_{k=1}(a_k+b_k)^2} \leq \sqrt{\sum^{n+1}_{k=1}(a_k)^2} + \sqrt{\sum^{n+1}_{k=1}(b_k)^2}$

Then $\sqrt{\sum^{n+1}_{k=1}(a_k+b_k)^2} = \sqrt{\sum^{n}_{k=1}(a_k+b_k)^2 +(a_{n+1}+ b_{n+1})^2} \leq \sqrt{\sum^{n}_{k=1}(a_k+b_k)^2}+ \sqrt{(a_{n+1}+ b_{n+1})^2} \leq \sqrt{\sum^n_{k=1}(a_k)^2} + \sqrt{\sum^n_{k=1}(b_k)^2}+(a_{n+1}+ b_{n+1})$

I am stuck here.

Answer 1

It's just a triangle inequality, but also we can use C-S: $$\sqrt{\sum_{k=1}^na_k^2}+\sqrt{\sum_{k=1}^nb_k^2}=\sqrt{\sum_{k=1}^n(a_k^2+b_k^2)+2\sqrt{\sum_{k=1}^na_k^2\sum_{k=1}^nb_k^2}}\geq$$ $$\geq\sqrt{\sum_{k=1}^n(a_k^2+b_k^2)+2\sum_{k=1}^na_kb_k}=\sqrt{\sum_{k=1}^n(a_k+b_k)^2}.$$

Answer 2

Using the CAUCHY-SCHWARZ inequiality: Let $x$ and $y$ vectors in $\mathbb{R}^n$. The Cauchy-Schwarz inequality states that $$|x\cdot y|\leq |x||y|$$ Written out in coordinates, this says $$|x_1y_1+x_2y_2+\dots+x_ny_n|\leq \sqrt{x_1^2+x_2^2+\dots+x_n^2}\sqrt{y_1^2+y_2^2+\dots+y_n^2}$$ So the Cauchy-Schwarz inequality tells us that $$ \begin{array}{ccl} |x+y|^2&=&(x+y)\cdot (x+y)\\ &=&|x|^2+|y|^2+2(x\cdot y)\\ &\leq& |x|^2+|y|^2+2|x||y|\\ &=&(|x|+|y|)^2 \end{array} $$ in other words $$\sqrt{\sum_{k=1}^n(x_k+y_k)^2}\leq \sqrt{\sum_{k=1}^n{x_k}^2}+\sqrt{\sum_{k=1}^n{y_k}^2}$$

Answer 3

Using inequality holder:here $$\sum_{i=1}^{n}(x_{i})(y_{i}) \leq(\sum_{i=1}^{n}(x_{i})^{p})^{\frac{1}{p}}(\sum_{i=1}^{n}(y_{i})^{q})^{\frac{1}{q}}$$
and we put $$\frac{1}{p}+\frac{1}{q}=1$$ where$$p\succ1,q\succ1$$and $$z_{i}=(x_{i}+y_{i})^{p-1}$$
$$\sum_{i=1}^{n}(x_{i})(z_{i})\leq(\sum_{i=1}^{n}(x_{i})^{p})^{\frac{1}{p}}(\sum_{i=1}^{n}(z_{i})^{q})^{\frac{1}{q}} \implies \sum_{i=1}^{n}(x_{i})(z_{i})\leq(\sum_{i=1}^{n}(x_{i})^{p}))^{\frac{1}{p}}(\sum_{i=1}^{n}(z_{i})^{\frac{p}{p-1}})^{\frac{p-1}{p}}$$

We have $$\frac{1}{p}+\frac{1}{q}=1\implies q=\frac{p}{p-1}$$

As well $(2)$ we find : $$\sum_{i=1}^{n}(y_{i})(z_{i})\leq(\sum_{i=1}^{n}(y_{i})^{p})^{\frac{1}{p}}(\sum_{i=1}^{n}(z_{i})^{\frac{p}{p-1}})^{\frac{p}{p-1}}$$\

Collects(1).(2) we find :\

$$\sum_{i=1}^{n}((x_{i})+(y_{i}))^{p}\leq\{(\sum_{i=1}^{n}(x_{i})^{p})^{\frac{1}{p}}+(\sum_{i=1}^{n}(y_{i})^{p})^{\frac{1}{p}}\}(\sum_{i=1}^{n}(z_{i})^{\frac{p}{p-1}})^{\frac{p-1}{p}} \implies(\sum_{i=1}^{n}((x_{i})+(y_{i}))^{p})^{\frac{1}{p}}\leq(\sum_{i=1}^{n}(x_{i})^{p})^{\frac{1}{p}}+(\sum_{i=1}^{n}(y_{i})^{p})^{\frac{1}{p}}$$

So for $p=2$ we fin this inequality .
$$(\sum_{i=1}^{n}(x_{i}+(y_{i})^{2})^{\frac{1}{2}}\leq(\sum_{i=1}^{n}(x_{i}^{2}))^{\frac{1}{2}}(\sum_{i=1}^{n}y_{i}^{2})^{\frac{1}{2}}$$