Question

If a, b and c are positive real numbers such that a + b + c = 3 then prove that :

\(\sqrt{a^3 + 3b} + \sqrt{b^3+ 3c} +\sqrt{c^3 +3a} \ge 6\)

Answer 1

Hello, Kush,
Great to hear from you!

Regarding the math problem you've posted: I don't immediately see the way to a complete proof. I do have a couple of suggestions to start with, however:

1) Experiment with actual values of a, b and c such that a+b+c=3. It goes without saying that a=1, b=1, c=1 is one such possibility. Substitute these three values back into the original equation. Is the resulting equation true or false? What, if anything, can you learn from this?

2) Experiment with eliminating one variable: for example, c-3-a-b.

3) Try eliminating the square roots by squaring both sides of the equation. You could, for example, take the original equation, move one of the three square roots to ther right side, and then square each of the (binomial) sides of the resulting equation. This would produce another equation that still has one or more square root operators, but you could eliminate it / them in the same manner.

Best to you!

Answer 2

Sir, thanks a lot for those suggestions.

1) for a = b = c = 1, I get that original inequality true with equality sign. That is, the LHS equals 6. The resulting equation is true. I'm not sure if I will be able to learn anything from this. What I think is that a , b and c should be greater than or equal to 1 for the inequality to hold.

2) Eliminating a variable (taking your example into consideration) gives :

\(c = 3 - (a+b)\)

\(\sqrt{a^3 + 3b} + \sqrt{b^3 + 3(3 - (a+b))} + \sqrt{(3-(a+b))^3 + 3a} \ge 6 \\
\sqrt{a^3+ 3b} + \sqrt{b^3 + 9 - 3a - 3b} +\sqrt{(3-(a+b))^3 + 3a} \ge 6\)

While if I keep simplifying it I don't get anything useful. Should I solve it further?

3) \(\sqrt{a^3 + 3b} + \sqrt{b^3 + 3c} \ge 6 - \sqrt{c^3 + 3a} \\
\text{Squaring Both Sides} \\
a^3 + 3b + b^3 + 3c + 2\sqrt{(a^3 + 3b)(b^3 + 3)} \ge 36 + c^3 + 3a - 12\sqrt{c^3 + 3a} \)

Answer 3

I tried to follow your guidelines, but there is nothing satisfactory yet. Though, I usually don't give up so earlier, but, I'm not finding any proper way to go on from the equations I got. As the square roots will be irritating later.

\((a^3 + b^3 + 3(b +c) ) + 2\sqrt{(a^3 + 3b)(b^3 +3c)} \ge 36 + c^3 + 3a - 12\sqrt{c^3 +3a} \\
2\sqrt{(a^3 + 3b)(b^3 +3c)} \ge 36 + c^3 - (a^3 + b^3 + 3(b+c)) + 3a - 12\sqrt{c^3 + 3a} \\
2\sqrt{(a^3 + 3b)(b^3 +3c)} + 12\sqrt{c^3 + 3a} \ge 36 + c^3 - a^3 - b^3 - 3(b+c) \)

Again, if I will square both sides, it will become more complex. :(

Answer 4

Challenging problem, yes. As you can see, I've approached it by identifying and considering a variety of approaches, since I've not encountered such a proof problem before. At this point I'd suggest you try brainstorming on other possible approaches and then try them out.

I understand you live in a relatively isolated place, but am wondering whether you have a colleague (fellow student) with whom you could discuss this and other problems.

Kush, i need to get off the 'Net soon. I hope to exchange further messages with you very soon. Best to you.

Answer 5

I will try to discuss this with my brother. Sure Sir. I will let you know if I get any method to solve this. Again, thanks a lot for your responses.

Answer 6

somehow i got the weaker bound for it ...it's equal to \( \sqrt {12} \)

Answer 7

What is equal to \(\sqrt{12}\) ? The LHS?

Answer 8

RHS.

Answer 9

Oh, wait, which equation are you referring to?

Answer 10

\[ \sqrt{a^3 + 3b} + \sqrt{b^3+ 3c} +\sqrt{c^3 +3a} \ge \sqrt{12} \]

Answer 11

Oh, okay, so you have proved this for \(\ge 2\sqrt{3}\) while we had to prove for \(\ge 6\) .
Though, this doesn't give us the required proof, but I will love to see how you proved that,

Answer 12

use the fact that
\[ a+b+c \ge \sqrt{a^2 + b^2 + c^2}\]
and
\[ a^3 + b^3+c^3 \ge 3 , a+b+c=3\]

Answer 13

there is long way to prove such inequalities using calculus, it called the method of lagrange multiplies ... but it's not nice and way too long.

Answer 14

\(3 \ge \sqrt{a^2 + b^2 + c^2}\)

and

\(a^3 + b^3 + c^3 \ge \sqrt{a^2 + b^2 + c^2}\)

I have not studied Lagranges' Multiples yet, and if it is way too long, then let us prefer to find a shorter method.

Answer 15

no ... make subsitution
\[ \sum \sqrt{a^3 + 3b} \ge \sqrt{\sum a^3 + 3b }\]

Answer 16

Okay, but, how did you get to that substitution ?

Answer 17

let x =a^3+3b

Answer 18

this is much more weaker than the inequality you state, it's useless to do it.

Answer 19

Yeah..

Answer 20

\( 9 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \ \
a^2 + b^2 + c^2 = 9 - 2(ab + bc + ca) \\
a^3 + b^3 +c^3 \ge \sqrt{a^2 + b^2 + c^2}\\
\implies a^3 + b^3 + c^3 \ge \sqrt{9 - 2(ab + bc + ca) } \\
\implies (a^6 + b^6 + c^6 + 2(a^3 b^3 + b^3c^3 + c^3a^3) \ge 9 -2ab -2bc - 2ac \\
\implies a^6 + b^6 + c^6 + 2ab(a^2b^2 + 1) + 2bc(b^2c^2 +1) + 2ac(c^2a^2 +1) \ge 9\)

While, I'm thinking to do something with this ^ :P not sure whether this will give us anything useful or not.

Answer 21

While at the other part of my mind, I am thinking to prove this :

\(\sqrt{a^3 + 3b} \ge 2\) ; \(\sqrt{b^3 +3c} \ge 2\) ; \(\sqrt{c^3 +3a} \ge 2\)

This will be much easier for us.

Let us just take one equation into consideration.

\(\boxed{\sqrt{a^3 +3b} \ge 2}\)

\(a^3 +3b \ge 4\)

This should be true.. hmm!

Answer 22

no that's incorrect ..

Answer 23

Try using Cauchy Schwarz inequality

Answer 24

just choose a=b=0, c=3

Answer 25

Oh.. okay, yep, I used wrong method.

Answer 26

@vishweshshrimali5 - can you explain it a litle bit, like how will it help us?

Answer 27

First of all do you know about it ?

Answer 28

http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

Answer 29

it seemed like you can find upper bound using cauchy swarz.

Answer 30

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpa1/t1.0-9/10371689_10152501448231552_8766121181245523511_n.jpg

Here is from where I got this identity. I don't know what has the user done there, whether he is proving this identity there or is doing something else.

Answer 31

He has used CS theorem.

Answer 32

Cauchy Schwarz inequality = CS ineq.

Answer 33

Oh, so he has proved it somewhere ? :P

Answer 34

Well, to be honest, I don't understand the CS inequality. Is this the only way to prove this identity? Well, I'm still trying to understand it...

Answer 35

It basically says that:

\[\large{(xa+yb)^2 \le (x^2 + y^2)(a^2 + b^2)}\]

Answer 36

As for the proof, search on internet.

Answer 37

No he has not proved this ineq. anywhere, it is a very common ineq. among olympiad students. So, he has taken it that the student knows it.

Answer 38

This will give you the very basic idea of the ineq.:
\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
It basically says that:

\[\large{(xa+yb)^2 \le (x^2 + y^2)(a^2 + b^2)}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 39

Oh, I see. I will try hard to understand this and get through the problem. Will post the solution I get.

@experimentX , @mathmale and @vishweshshrimali5 - You guys are awesome. Thanks a lot Everyone.

Answer 40

np at all. And others did the most work, I just gave one single hint. :)

Answer 41

http://www.aam.org.in/site/st_material/14.pdf

Answer 42

@experimentX - That pdf is really nice. I would love to read it :-)

Thanks.