# given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$

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

• $uvw$ kills it as its lenear in $w^3$.... – Albus Dumbledore Feb 20 at 8:32
• I didn't understand, I don't know any high level theorms, pl tell me what u mean In detail – Mehul Feb 20 at 8:34
• You can use AM-GM here, like my solution. – tthnew Feb 20 at 9:31

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!

• Wow, MV method! Can you check my solution? – tthnew Feb 20 at 9:28
• @tthnew your is also nice ,didnt think it was possible by AM-GM (+1) – Albus Dumbledore Feb 20 at 9:33
• Thanks, +1 to you also. – tthnew Feb 20 at 9:33
• Nice solution indeed! – Math Lover Feb 20 at 9:48
• 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 $a=b=c$. – Martin R Feb 20 at 10:33

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 $$(ab-bc)^2+(bc-ca)^2+(ca-ab)^2 \geq 0$$.

• What do you mean for substituting in (i) ? – tthnew Feb 20 at 9:33
• +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 $p=a+b+c,q=ab+bc+ca,r=abc$ :) – Albus Dumbledore Feb 20 at 9:53

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!