Prove this inequality

If a,ba,b and cc are distinct positive numbers, prove that a2+b2+c2>(a+b+c)24a^2 + b^2 + c^2 > \dfrac {(a+b+c)^2}4 .

#Algebra

Note by Monishwaran Maheswaran
4 years, 11 months ago

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Comments

Just a direct application of Titu's lemma.

a21+b21+c21>(a+b+c)23\dfrac{a^2}1+\dfrac{b^2}1+\dfrac{c^2}1>\dfrac{(a+b+c)^2}{3}

Now since (a+b+c)23>(a+b+c)24\dfrac{(a+b+c)^2}{3}>\dfrac{(a+b+c)^2}{4} (Obviously since 3<43<4).

Hence we get:

a2+b2+c2>(a+b+c)24a^2+b^2+c^2>\dfrac{(a+b+c)^2}{4}

Rishabh Jain - 4 years, 11 months ago

By Power mean inequality (QAGH), QM(a,b,c)>AM(a,b,c)⇒a2+b2+c23>a+b+c3⇒a2+b2+c2>(a+b+c)23>(a+b+c)24  .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 \; .

Pi Han Goh - 4 years, 11 months ago

a2+b2+c2>(a+b+c)244(a2+b2+c2)>(a+b+c)2  ⟹  3(a2+b2+c2)>2(ab+ac+bc)a2+b2+c2>ab+ac+bc→(∗)  ⟹  3(a2+b2+c2)>2(a2+b2+c2)>2(ab+ac+bc)3(a2+b2+c2)>2(ab+ac+bc)Hence proveda^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}

(∗)  ⟹  a2+b2+c2≥ab+ac+bcThis is provable using a variety of ways,using Rearrangement, AM-GM,Algebraic Manipulation etc.I’ll show a proof using Algebraic Manipulation2a2+2b2+2c2≥2ab+2bc+2ac  ⟹  (a2−2ab+b2)+(b2−2ac+c2)+(c2−2ac+a2)≥0(a−b)2+(b−c)2+(c−a)2≥0Equality occurs when  a=b=c  However,since a,b,c are distinct,this precludes equality,hence  a2+b2+c2>ab+ac+bc(*)\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

Abdur Rehman Zahid - 4 years, 11 months ago
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