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# 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 $$abc=1$$, 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.

Note by Figel Ilham
1 year, 8 months ago

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You can simply use cauchy schwarz inequality · 1 year, 8 months ago