Prove that $(\Sigma a^2)^5\ge27~(\Sigma a^3b^2)^2$

Question

$\newcommand{\cyc}{\sum\limits_{\rm cyc}}$You can use \cyc for $\cyc$ in this topic.

Let $a$, $b$, $c$ be positive real numbers. Prove that \[\left(a^2+b^2+c^2\right)^5\ge27\cdot\left(a^3b^2+b^3c^2+c^3a^2\right)^2.\]

I used Vasc inequality for symmetry, and did \[27\left(\cyc a^3b^2\right)^2\le27\left(\cyc a^3b\right)\cyc a^3b^3\le9\left(\cyc a^2\right)^2\cyc a^3b^3,\] so it suffices to show that $9\cyc a^3b^3\le\left(\cyc a^2\right)^3$, which is wrong.

Anyway, here is a computer proof (which might hurt your eyes):

$$\tiny\frac{11133081393065191748404452685704727703842711321128011221788442356680779389}{83325406159945225773761215993180645052139660969167854665096869468319128387}\sum a^{8} \left(a - b\right) \left(a - c\right)+ \frac{75774499776880396841723729613601191138243621095597037566722116653917854393}{166650812319890451547522431986361290104279321938335709330193738936638256774}\sum c^{8} \left(- a + c\right)^{2}+ \frac{287323656211716366782629529381247989009342413166649177769899908948317724874}{416627030799726128868806079965903225260698304845839273325484347341595641935}\sum a c^{7} \left(- a + c\right)^{2}+ \frac{1391554600294356965313066277427777840564943689822584084547601231259917407499}{166650812319890451547522431986361290104279321938335709330193738936638256774}\sum b^{2} c^{2} \left(b - c\right)^{2} \left(- a c + b^{2}\right)^{2}\\\tiny + \frac{1}{166650812319890451547522431986361290104279321938335709330193738936638256774}\sum \left(- a c^{2} + b^{3}\right)^{2} \cdot \left(15750846263137681648990801621787501156439960918142340787660435274253162358 a^{2} b c + 92604232176822259801347585053143855872107668854554129225460921629790186727 a b^{3}\right)+ \frac{1}{166650812319890451547522431986361290104279321938335709330193738936638256774}\sum \left(- a^{2} c + b^{3}\right)^{2} \cdot \left(683645438235834405209905496856140088914204369410088364043747433096154783078 a^{2} b c + 133019011894066117189213768481719794867984956592149394818583897365959382087 a b^{3}\right)\\ \tiny+ \frac{282475249}{880082939861341474622465963319973973040699099156350880972753135324386634023494}\sum \left(90574269046915887451745582632378457112210906624555105334876926486355 c^{2} \left(- a + c\right)^{2} + 99230852260528550529633100797542068662065716336949040650494569065194 \left(b - c\right)^{4}\right) \left(- a + \frac{64 b}{49} - \frac{15 c}{49}\right)^{6}\\\tiny + \frac{2401}{833254061599452257737612159931806450521396609691678546650968694683191283870}\sum \left(- a + \frac{64 b}{49} - \frac{15 c}{49}\right)^{2} \cdot \left(5409681167184077883380073923421695011549107239679735001724818283813225033 a^{5} b^{3} + 2921312415349208904554555591957347275658143812840288512751913363993961982 a^{4} b^{4} + 1377465103746322591875353559649701871922710313072499388318363884491780622 a^{3} b^{5} + 439922828548523903587716170365365775897773071922008685602782360760643305 a^{2} b^{6}\right)\\ \tiny+ \frac{117649}{49708496976487043939743646666183258555565415442404843700505058351903552730194548827870}\sum \left(- a + \frac{64 b}{49} - \frac{15 c}{49}\right)^{4} \cdot \left(257398984898387109858400197684299324300071518550592275725667530649767177106514141 a^{5} b + 25807946631013965492968426826705783363949464129171593632428864541127682246815367 a^{3} b \left(a - b\right)^{2} + 126415391863922213317307374969629596885161578539202372210119193641330456244003960 a^{3} c \left(- a + c\right)^{2} + 37955200021455815083032605127392251002903819164489134730609405681888253558872385 a^{2} b^{2} c^{2} + 241558948026036964720337597588301580736096741734501258610066400510824391244408185 a^{2} \left(- a + c\right)^{2} \left(b - c\right)^{2} + 331091243393787448964139893711547017132956156761319187250481783478264425185929190 b^{3} c \left(b - c\right)^{2} + 36695686941502885379190142959806664491718484608190012210539716689766114510936190 \left(b - c\right)^{6}\right).$$

I'd like to see a proof by hand!

Answer 1

Hint: As the inequality is homogeneous, we may set $a^2+b^2+c^2=3$ and instead prove equivalently that $\sum_{cyc} a^3b^2\leqslant 3$, which can be shown by Rearrangement Inequality.

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Here is an alternate way using Vasc's inequality (and others). First, with CS we have $$\left(\sum a^3b^2\right)^2 \leqslant \left(\sum a^2b^2\right)\cdot \left(\sum a^4b^2\right)$$ Hence it is enough to show for positives $x, y, z$ s.t. $x+y+z=3$, $$\left(\sum xy\right) \cdot \left(\sum x^2y\right)\leqslant 9 \tag{1}$$ Note all the $\sum$ refer to cyclic sums and let $s=\sum xy \in (0, 3]$ for convenience. Further, we have $$3\sum x^2y = \left(\sum x\right)\cdot \left(\sum x^2y\right)=\sum x^3y + \sum x^2y^2+3xyz \tag{2}$$ We now bound the RHS terms like so: $$3\sum x^3y \leqslant \left(\sum x^2\right)^2=(9-2s)^2\tag{Vasc}$$ $$\sum x^2y^2 = \left(\sum xy\right)^2-6xyz$$ $$3xyz \geqslant 4\sum xy-9=4s-9 \tag{Schur}$$ Using the above three in $(2$), we get $$9\sum x^2y\leqslant (9-2s)^2+3(s^2-6xyz)+9xyz\leqslant (9-2s)^2+3s^2-3(4s-9)\\=7s^2-48s+108$$

Hence to prove $(1)$, it is enough to show $$s\cdot (7s^2-48s+108)\leqslant 81$$ $$\iff (3-s)(7s^2-27s+27)\geqslant 0$$ which is obvious.

Answer 2

It's my very old problem: https://artofproblemsolving.com/community/c6h72296

Since $(a,b,c)$ and $(a^2b^2,a^2c^2,b^2c^2)$ have the same ordering, by Rearrangement and AM-GM twice we obtain: $$\sum_{cyc}a^3b^2=a\cdot a^2b^2+b\cdot b^2c^2+c\cdot c^2a^2\leq$$ $$\leq a\cdot a^2b^2+b\cdot a^2c^2+c\cdot b^2c^2=b(a^3b+a^2c^2+bc^3)=$$ $$=b\left(a^2\left(ab+\frac{c^2}{2}\right)+c^2\left(\frac{a^2}{2}+bc\right)\right)\leq$$ $$\leq b\left(a^2\left(\frac{a^2+b^2}{2}+\frac{c^2}{2}\right)+c^2\left(\frac{a^2}{2}+\frac{b^2+c^2}{2}\right)\right)=$$ $$=\sqrt{\frac{b^2(a^2+c^2)^2}{4}}(a^2+b^2+c^2)\leq\sqrt{\left(\frac{b^2+2\cdot\frac{a^2+c^2}{2}}{3}\right)^3}(a^2+b^2+c^2)=$$ $$=\sqrt{\frac{(a^2+b^2+c^2)^5}{27}}.$$