How do you prove that sqrt(a+b+c) <= sqrt(a)+sqrt(b)+sqrt(c) given that a,b,c >=0

X049

New member
Joined
May 26, 2021
Messages
8
I am not asking for the answer, just someone to guide me in the right direction
 
Last edited:
One way would be to prove by induction that

[MATH]n \in \mathbb Z, \ i \in \mathbb Z^+ \text { and } 1 \le i \le n + 1, \text { and } r_i \ge 0 \implies \sqrt{\sum_{i=1}^{n+1}r_i} \le \left ( \sum_{i=1}^{n+1} \sqrt{r_i} \right ).[/MATH]
[MATH]\text {Assume } n = 1 \text { let } x = \sqrt{\sum_{i=1}^{n+1} r_i} = \sqrt{r_1 + r_2}, \text { and let } y = \left ( \sum_{i=1}^{n+1} \sqrt{r_i} \right ) = \sqrt{r_1} + \sqrt{r_2}.[/MATH]
[MATH]\therefore x^2 = r_1 + r_2 \text { and } y^2 = r_1 + 2\sqrt{r_1r_2} + r_2.[/MATH]
Now what?
 
Consider as an approach that you could let \(\displaystyle \ a = x^2, \ b = y^2, \ c = z^2.\)

In general, \(\displaystyle \ \sqrt{M} = |M|,\ \) but it equals M if M is greater than or equal to zero.

And note that here a, b, c are nonnegative.

Substituting in the variables we have:

\(\displaystyle \sqrt{x^2 + y^2 + z^2} \ \ vs. \ \sqrt{x^2} + \ \sqrt{y^2} + \sqrt{z^2} \ \implies \)

\(\displaystyle \sqrt{x^2 + y^2 + z^2} \ \ vs. \ \ x + y + z \ \implies ?\)

Now, square each side and compare them to be able to confirm the direction of
the inequality symbol for the proof.
 
We want to prove: [MATH]\sqrt{a}+\sqrt{b}+\sqrt{c}≥\sqrt{a+b+c} \hspace2ex[/MATH] (1)
Note [MATH]a, b, c ≥0[/MATH], so they can be written as [MATH]a=x^2, b=y^2, c=z^2[/MATH]
Also, we can choose [MATH]x, y, z[/MATH] to be: [MATH]x≥0, y≥0, z≥0[/MATH]Now, e.g. [MATH]\sqrt{a}=\sqrt{x^2}=x \hspace1ex[/MATH] since [MATH]x≥0[/MATH]
So expression (1) becomes: [MATH]x+y+z≥\sqrt{x^2+y^2+z^2}[/MATH]Since all terms are [MATH]≥0[/MATH], this inequality holds:
[MATH]\Leftrightarrow (x+y+z)^2≥x^2+y^2+z^2[/MATH][MATH]\Leftrightarrow (x+y+z)^2-(x^2+y^2+z^2)≥0 \hspace2ex[/MATH] (2)
When you expand and simplify this expression, you should be able to use the fact that [MATH]x≥0, y≥0, z≥0[/MATH] to show that it is [MATH]≥0[/MATH].
 
Direct proof:
\(\displaystyle (\sqrt{a}+ \sqrt{b}+ \sqrt{c})^2= \sqrt{a}(\sqrt{a}+ \sqrt{b}+ \sqrt{c})+ \sqrt{b}(\sqrt{a}+ \sqrt{b}+ \sqrt{c})+ \sqrt{c}(\sqrt{a}+ \sqrt{b}+ \sqrt{c})\)
\(\displaystyle = (\sqrt{a}\sqrt{a}+ \sqrt{a}\sqrt{b}+ \sqrt{a}\sqrt{c})+ (\sqrt{b}\sqrt{a}+ \sqrt{b}\sqrt{b}+ \sqrt{b}\sqrt{c})+ (\sqrt{c}\sqrt{a}+ \sqrt{c}\sqrt{b}+ \sqrt{c}\sqrt{c})\)
\(\displaystyle = a+ b+ c+ 2\sqrt{ab}+ 2\sqrt{ac}+ 2\sqrt{bc}\).

\(\displaystyle (\sqrt{a}+ \sqrt{b}+ \sqrt{c})^2\) is a+b+ c plus other non-negative numbers.
\(\displaystyle (\sqrt{a}+ \sqrt{b}+ \sqrt{c})^2\) is larger than a+ b+ c so, since everything is non-negative, taking the square root of both sidess, \(\displaystyle \sqrt{a}+ \sqrt{b}+ \sqrt{c}\ge \sqrt{a+ b+ c}\)
 
  • Like
Reactions: lex
Top