# Prove this inequality

If $$a,b$$ and $$c$$ are distinct positive numbers, prove that $$a^2 + b^2 + c^2 > \dfrac {(a+b+c)^2}4$$.

Note by Monishwaran Maheswaran
2 years ago

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$a^2+b^2+c^2>\frac{(a+b+c)^2}{4}\\ 4(a^2+b^2+c^2)>(a+b+c)^2\\ \implies 3(a^2+b^2+c^2)>2(ab+ac+bc)\\ \boxed{a^2+b^2+c^2>ab+ac+bc\rightarrow (*)}\implies 3(a^2+b^2+c^2)>2(a^2+b^2+c^2)>2(ab+ac+bc)\\ 3(a^2+b^2+c^2)>2(ab+ac+bc)\\ \text{Hence proved}$

$(*)\implies a^2+b^2+c^2≥ab+ac+bc\\ \text{This is provable using a variety of ways,using Rearrangement, AM-GM,Algebraic Manipulation etc.I'll show a proof using Algebraic Manipulation}\\ 2a^2+2b^2+2c^2\geq 2ab+2bc+2ac\\ \implies (a^2-2ab+b^2)+(b^2-2ac+c^2)+(c^2-2ac+a^2)\geq 0\\ \boxed{(a-b)^2+(b-c)^2+(c-a)^2\geq 0}\\ \text{Equality occurs when}\; a=b=c\;\text{However,since a,b,c are distinct,this precludes equality,hence}\; a^2+b^2+c^2>ab+ac+bc$

- 2 years ago

By Power mean inequality (QAGH), $QM(a,b,c) > AM(a,b,c) \Rightarrow \sqrt{ \dfrac{a^2+ b^2+c^2}3 } > \dfrac{a+b+c}3 \Rightarrow a^2 + b^2 + c^2 > \dfrac{(a+b+c)^2}3 > \dfrac{(a+b+c)^2}4 \; .$

- 2 years ago

Just a direct application of Titu's lemma.

$\dfrac{a^2}1+\dfrac{b^2}1+\dfrac{c^2}1>\dfrac{(a+b+c)^2}{3}$

Now since $$\dfrac{(a+b+c)^2}{3}>\dfrac{(a+b+c)^2}{4}$$ (Obviously since $$3<4$$).

Hence we get:

$a^2+b^2+c^2>\dfrac{(a+b+c)^2}{4}$

- 2 years ago