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Let $a, b, c$ are non-negative real numbers satisfy $a+b+c=3$. Prove that $$\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2\geq\frac{3}{4}$$

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    i haven't idea for this post, help me ! – 2017-02-24
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    Have you copied the question wrong? Are there more conditions on $a,b,c$? – 2017-02-24
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    oh, i'm sorry because my foolishness – 2017-02-24
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    So is the question you typed here the correct one? – 2017-02-24
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    and what is the correct Statement? – 2017-02-24
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    sorry i don't know fix where – 2017-02-24

6 Answers 6

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It's wrong! Try please $a=b=c=1$.

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    you can fix it ? – 2017-02-24
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    @Word Shallow What do you mean? – 2017-02-24
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    you can fix my question ? – 2017-02-24
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    @Word Shallow Tell my, what I need to write and I'll fix. – 2017-02-24
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    @Word Shallow I found the following similar question. Let $a$, $b$ and $c$ be real numbers such that $a+b+c=3$. Prove that: $\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2\geq\frac{3}{4}$ – 2017-02-24
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Using the fact that $RMS\geq A.M$

$\sqrt{\dfrac{\left(a-\dfrac{1}{2}\right)^2+\left(b-\dfrac{1}{2}\right)^2+\left(c-\dfrac{1}{2}\right)^2}{3}}\geq \dfrac{\left(a-\dfrac{1}{2}\right)+\left(b-\dfrac{1}{2}\right)+\left(c-\dfrac{1}{2}\right)}{3}$

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\begin{align*} \left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2 &\ge \frac{\left(\left|a-\frac{1}{2} \right| +\left|b-\frac{1}{2} \right|+\left|c-\frac{1}{2} \right|\right)^2}{3}\\ &\ge \frac{\left(\left|a-\frac{1}{2} + b-\frac{1}{2}+c-\frac{1}{2} \right|\right)^2}{3}\\ &= \frac{3}{4}. \end{align*}

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Cauchy Schwarz:

$$u+v+w = (1,1,1) \cdot (u,v,w) \le \sqrt{3} \sqrt{ u^2+v^2 + w^2},$$

with equality if and only if $u=v=w\ge 0$.

Set $u=a-\frac12,v=b-\frac 12,w=c-\frac 12$ and square both sides to obtain

$$\frac 94 \le 3 \left ( (a-\frac 12)^2 + (b-\frac 12)^2 + (c-\frac 12)^2 \right)$$

Divid both sides by $3$ to obtain the result, with equality if and only if $a=b=c=1$.

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Note that, for any real numbers $a$, $b$, and $c$, we have

$$\begin{align} 0&\le(a-b)^2+(b-c)^2+(c-a)^2\\ &=2(a^2+b^2+c^2-ab-bc-ca)\\ &=3(a^2+b^2+c^2)-(a+b+c)^2 \end{align}$$

Now if $a+b+c=3$, then $0\le3(a^2+b^2+c^2)-3(a+b+c)$, which, on dropping the $3$ and adding three $1\over4$'s on both sides, leads to

$$\begin{align} {3\over4}&\le\left(a^2-a+{1\over4}\right)+\left(b^2-b+{1\over4}\right)+\left(c^2-c+{1\over4}\right)\\ &=\left(a-{1\over2}\right)^2+\left(b-{1\over2}\right)^2+\left(c-{1\over2}\right)^2 \end{align}$$

The assumption that $a$, $b$ and $c$ are non-negative is unnecessary.

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Hint: this question includes strong geometric interpretation: $a+b+c=3$ represents a plane in 3D space, and inequality is a presentation of a sphere. Calculate the distance from the plane to the centre of sphere.