AM-GM Struggle! (1)

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

For \(a,b,c \in \mathbb{R^{+}}\), if \(abc=1\), then proof that \[a^2+b^2+c^2 \geq a+b+c\]

My proof is like this:

Since abc=1abc=1, then according to AM-GM, we provide a+b+c3≥abc3\frac{a+b+c}{3} \geq \sqrt[3]{abc} ⇒a+b+c≥3\Rightarrow a+b+c \geq 3

Consider these: (2a−1)2+(2b−1)2+(2c−1)2≥0(2a-1)^2 + (2b-1)^2 + (2c-1)^2 \geq 0

by QM-AM, we provide (2a−1)2+(2b−1)2+(2c−1)23≥(2a−1)+(2b−1)+(2c−1)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 (2a−1)2+(2b−1)2+(2c−1)23≥2(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. (2a−1)2+(2b−1)2+(2c−1)23≥1\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq 1 ⇒(2a−1)2+(2b−1)2+(2c−1)2≥3\Rightarrow (2a-1)^2+(2b-1)^2+(2c-1)^2 \geq 3 ⇒4(a2+b2+c2)−4(a+b+c)+3≥3\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+c2≥a+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 (2a−1)2+(2b−1)2+(2c−1)2≥3(2a-1)^2 +(2b-1)^2 + (2c-1)^2 \geq 3 Is that valid for the proof? Any comments will be appreciated.

#Algebra #Inequalities #NeedHelp #Struggle

Note by Figel Ilham
6 years, 1 month ago

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Comments

You can simply use cauchy schwarz inequality

Matteo De Zorzi - 6 years, 1 month ago

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

Figel Ilham - 6 years, 1 month ago
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