Cauchy-Shwarz Inequality is:

$$(a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$$

However, it can be manipulated as:

$$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2}$$

I'm tasked with proving the following inequality:

$$\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + (a_2 + b_2 + \cdots + z_2)^2 + \cdots + (a_n + b_n \cdots + z_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + z_2^2 + \cdots + z_n^2}$$

I've proved both Cauchy-Shwarz and its manipulation, but am lost when it come to the inequality right above. Hints and/or solutions are welcome.

  • $\begingroup$ Can't you just replace $b_1$ by $-b_1$ in your manipulated version and then use induction? $\endgroup$ – Michael Burr Jan 4 '17 at 22:42
  • $\begingroup$ if you change $b_1$ by $-z_1$ in manipulated version wont it be enough? What do three dots mean in the left of the last inequality? $\endgroup$ – Markoff Chainz Jan 4 '17 at 22:43


Replacing $b_i$ by $-b_i$ one can transform

$$ \sqrt{(a_1-b_1)^2+\cdots+(a_n-b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2} $$ into $$ \sqrt{(a_1+b_1)^2+\cdots+(a_n+b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2} $$

Then, \begin{align*} \sqrt{(a_1+b_1+c_1)^2+\cdots+(a_n+b_n+c_n)^2}&=\sqrt{(a_1+(b_1+c_1))^2+\cdots+(a_n+(b_n+c_n))^2}\\ &\leq\sqrt{a_1^2+\cdots+a_n^2}+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}. \end{align*} Applying the result again, to the second radical, we get \begin{align*} \sqrt{a_1^2+\cdots+a_n^2}&+\sqrt{(b_1+c_1)^2+\cdots+(b_n+c_n)^2}\\ &\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}+\sqrt{c_1^2+\cdots+c_n^2}. \end{align*} Now, use induction to complete the proof.


Let $\vec{a}(a_1,a_2,...,a_n)$, $\vec{b}(b_1,b_2,...,b_n)$,...,$\vec{z}(z_1,z_2,...,z_n)$.

Hence, $|\vec{a}|+|\vec{b}|+...+|\vec{z}|\geq|\vec{a}+\vec{b}+...+\vec{z}|$ and we are done

because in our case we got $|\vec{a}|+|\vec{b}|\geq|\vec{a}+\vec{b}|$.


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