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

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

by QM-AM, we provide \[\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq \frac{(2a-1)+(2b-1) + (2c-1)}{3} = \frac{2(a+b+c)-3}{3}\]

Inserting the first AM-GM inequality, we get \[\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq \frac{2(a+b+c)-3}{3}\] \[\geq \frac{2(3)-3}{3} = 1\]

Then, solve it. \[\sqrt{\frac{(2a-1)^2 + (2b-1)^2 + (2c-1)^2}{3}} \geq 1\] \[\Rightarrow (2a-1)^2+(2b-1)^2+(2c-1)^2 \geq 3\] \[\Rightarrow 4(a^2+b^2+c^2) - 4(a+b+c) +3 \geq 3\] \[\Rightarrow 4(a^2+b^2+c^2)-4(a+b+c) \geq 0\] \[\Rightarrow a^2 +b^2 +c^2 \geq a+b+c\] \[ \blacksquare\]

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

## Comments

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TopNewestYou can simply use cauchy schwarz inequality – Matteo De Zorzi · 2 years, 2 months ago

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– Figel Ilham · 2 years, 2 months ago

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