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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

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

Log in to reply

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

Log in to reply

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:

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

Log in to reply