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I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...

Problem-

Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)

Developments-

Firstly applying am-gm on $a+b+c$ to get $abc\leq1$

Then applying quadratic mean arthematic mean to get

$a^2+b^2+c^2 \ge 3$

Then writing the expansion of $(a+b+c)^2$ we get

$a^2+b^2+c^2 = 9-2(ab+bc+ca)$ ...(I)

Thn applying am-hm on $ab+bc+ca$ to get

$ab+bc+ca \ge 3abc$

Then substituting in (I) we get

$a^2+b^2+c^2 \leq 9-6(abc)$

Then we substituting in (0)

$LHS \leq abc(9-6abc)$

Whose max value is $27/8$ but is not less than three, so I am stuck for this point onwards

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  • $\begingroup$ $uvw$ kills it as its lenear in $w^3$.... $\endgroup$ – Albus Dumbledore Feb 20 at 8:32
  • $\begingroup$ I didn't understand, I don't know any high level theorms, pl tell me what u mean In detail $\endgroup$ – Mehul Feb 20 at 8:34
  • $\begingroup$ You can use AM-GM here, like my solution. $\endgroup$ – tthnew Feb 20 at 9:31
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WLOG $a\le b\le c$ so $\frac{b+c}{2}\ge 1$

let $$f(a,b,c)= abc(a^2+b^2+c^2)$$ now $$f(a,b,c)-f(a,\frac{b+c}{2},\frac{b+c}{2})=-\frac{1}{8} a (b - c)^2 (2 a^2 + b^2 - 2 b c + c^2)\le 0$$ let $t=\frac{b+c}{2}\ge 1$ So it suffices to show $$f(a,t,t)\le 3 \iff f(3-2t,t,t)\le 3$$ but $$f(3-2t,t,t)-3=-3 (t - 1)^2 (4 t^3 - 6 t^2 + 2 t + 1)\le 0$$ which is true because $4t^3-6t^2+2t=2t(t-1)(2t-1)\ge 0$

Done!

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  • $\begingroup$ Wow, MV method! Can you check my solution? $\endgroup$ – tthnew Feb 20 at 9:28
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    $\begingroup$ @tthnew your is also nice ,didnt think it was possible by AM-GM (+1) $\endgroup$ – Albus Dumbledore Feb 20 at 9:33
  • $\begingroup$ Thanks, +1 to you also. $\endgroup$ – tthnew Feb 20 at 9:33
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    $\begingroup$ Nice solution indeed! $\endgroup$ – Math Lover Feb 20 at 9:48
  • $\begingroup$ Note that $f(a,b,c) \le f(a,\frac{b+c}{2},\frac{b+c}{2})$ together with the symmetry of $f$ and the domain already implies that the maximum is attained where $a=b=c$. $\endgroup$ – Martin R Feb 20 at 10:33
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I will use a well known inequality for my solution -

$a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$

$\implies (ab+bc+ca)^2 \geq 3abc(a+b+c) = 9abc$

Now using AM-GM,

$a^2+b^2+c^2+2 (ab + bc + ca) \geq 3 [(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$

$9 \geq 3[(a^2+b^2+c^2)(ab+bc+ca)^2]^{1/3}$

$3^3 \geq 9 abc(a^2+b^2+c^2)$

$3 \geq abc(a^2+b^2+c^2)$


On the inequality I used to answer comes from $(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$.

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  • $\begingroup$ What do you mean for substituting in (i) ? $\endgroup$ – tthnew Feb 20 at 9:33
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    $\begingroup$ +1 to be frank your solution is the best as it does not rely on crazy factorisation ,again,that substituiting in $(i)$ is confusing until second thought.I think your proof would be greatly simplified if you use the sub $p=a+b+c,q=ab+bc+ca,r=abc$ :) $\endgroup$ – Albus Dumbledore Feb 20 at 9:53
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Assume $a=\min\{a,b,c\} \rightarrow 0\le a \le 1.$ By AM-GM, we have $$abc\left(a^2+b^2+c^2\right) =\dfrac{1}{3} a\cdot 3bc\cdot\left(a^2+b^2+c^2\right)$$ $$\le \dfrac{1}{12} a \left[a^2+bc+\left(b+c\right)^2\right]^2\le \dfrac{1}{12} a \left[a^2+\dfrac{\left(3-a\right)^2}{4}+\left(3-a\right)^2\right]^2\le 3,$$ so it suffices to prove $$\left( 9\,{a}^{2}-33\,a+64 \right) \left(1-a \right) ^{3}\ge 0,$$ which is obvious since $0\le a\le 1.$

Done!

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