AM-GM Struggle! (1)

I have a problem, which is not really sure for me. Absolutely, it is about the inequalities!

For a,b,cR+a,b,c \in \mathbb{R^{+}}, if abc=1abc=1, then proof that a2+b2+c2a+b+ca^2+b^2+c^2 \geq a+b+c

My proof is like this:

Since abc=1abc=1, then according to AM-GM, we provide a+b+c3abc3\frac{a+b+c}{3} \geq \sqrt[3]{abc} a+b+c3\Rightarrow a+b+c \geq 3

Consider these: (2a1)2+(2b1)2+(2c1)20(2a-1)^2 + (2b-1)^2 + (2c-1)^2 \geq 0

by QM-AM, we provide (2a1)2+(2b1)2+(2c1)23(2a1)+(2b1)+(2c1)3=2(a+b+c)33\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq \frac{(2a-1)+(2b-1) + (2c-1)}{3} = \frac{2(a+b+c)-3}{3}

Inserting the first AM-GM inequality, we get (2a1)2+(2b1)2+(2c1)232(a+b+c)33\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq \frac{2(a+b+c)-3}{3} 2(3)33=1\geq \frac{2(3)-3}{3} = 1

Then, solve it. (2a1)2+(2b1)2+(2c1)231\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq 1 (2a1)2+(2b1)2+(2c1)23\Rightarrow (2a-1)^2+(2b-1)^2+(2c-1)^2 \geq 3 4(a2+b2+c2)4(a+b+c)+33\Rightarrow 4(a^2+b^2+c^2) - 4(a+b+c) +3 \geq 3 4(a2+b2+c2)4(a+b+c)0\Rightarrow 4(a^2+b^2+c^2)-4(a+b+c) \geq 0 a2+b2+c2a+b+c\Rightarrow a^2 +b^2 +c^2 \geq a+b+c \blacksquare

I had been inspired by filling the perfect squares so it provides (2a1)2+(2b1)2+(2c1)23(2a-1)^2 +(2b-1)^2 + (2c-1)^2 \geq 3 Is that valid for the proof? Any comments will be appreciated.

Note by Figel Ilham
4 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

You can simply use cauchy schwarz inequality

Matteo De Zorzi - 4 years, 5 months ago

Log in to reply

I got it. But, I want to learn advance for AM-GM topic.

Figel Ilham - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...