The inequality is equivalently with the inequality $x,y,z>0\Longrightarrow 6xyz\le \sum x(y^2+z^2)$ or with the inequality $R\ge 2r$.

I have a little bit ugly solution. Let $a=x+y, b=y+z,c=z+x$. Then we get $\sum_{sym} x^2 y \geq 6xyz$. Using $AM-GM$, $Q.E.D.$

Suppose $a,b,c$ are the sides of a triangle $ABC$ with inradius $r$. Prove that \[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \geq 12r^2(a+b+c). \] \[ a(b+c-a)^2+b(a+c-b)^2+c(a+b-c)^2 \geq 12r^2(a+b+c). \] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=506164&p=2842947#p2842947

$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc $ $ \Leftrightarrow $ $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}\leq 3 $(APMC2001)

xzlbq wrote: xzlbq wrote: xzlbq wrote: $4abcr(\frac{1}{h_b-2r+h_c}+\frac{1}{h_a-2r+h_c}+\frac{1}{h_a-2r+h_b})$ $ \geq a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)$ Note: $\frac{1}{4}\frac{a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)}{abc}\geq \sum_{cyc}{\frac{\sin{\frac{B}{2}}\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}}$ $3\geq \frac{4}{3}(\cos^2{\frac{C}{2}}+\cos^2{\frac{A}{2}}+\cos^2{\frac{B}{2}})$ $\geq 4(\frac{\tan{\frac{B}{2}}}{\cot{\frac{C}{2}}+\tan{\frac{B}{2}}}+\frac{\tan{\frac{C}{2}}}{\tan{\frac{C}{2}}+\cot{\frac{A}{2}}}+\frac{\tan{\frac{A}{2}}}{\tan{\frac{A}{2}}+\cot{\frac{B}{2}}})$ $\geq \frac{4r}{h_b-2r+h_c}+\frac{4r}{h_a-2r+h_c}+\frac{4r}{h_a-2r+h_b}$ $ \geq \frac{a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)}{abc}$ $3\geq 4(\frac{a}{2a+b+c}+\frac{b}{2b+c+a}+\frac{c}{b+2c+a})$ $\geq \frac{2}{3}(\frac{h_b+h_c}{h_b-2r+h_c}+\frac{h_c+h_a}{h_a-2r+h_c}+\frac{h_a+h_b}{h_a-2r+h_b})$ $\geq \frac{4}{3}(\cos^2{\frac{C}{2}}+\cos^2{\frac{A}{2}}+\cos^2{\frac{B}{2}})$ $\geq 4(\frac{\tan{\frac{B}{2}}}{\cot{\frac{C}{2}}+\tan{\frac{B}{2}}}+\frac{\tan{\frac{C}{2}}}{\tan{\frac{C}{2}}+\cot{\frac{A}{2}}}+\frac{\tan{\frac{A}{2}}}{\tan{\frac{A}{2}}+\cot{\frac{B}{2}}})$ $\geq \frac{4r}{h_b-2r+h_c}+\frac{4r}{h_a-2r+h_c}+\frac{4r}{h_a-2r+h_b}$ $ \geq \frac{a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)}{abc}$ $3\geq \frac{7}{2}(\frac{a}{s+c+a}+\frac{b}{s+a+b}+\frac{c}{s+b+c})$ $\geq 4(\frac{a}{2a+b+c}+\frac{b}{2b+c+a}+\frac{c}{b+2c+a})$ $\geq \frac{2}{3}(\frac{h_b+h_c}{h_b-2r+h_c}+\frac{h_c+h_a}{h_a-2r+h_c}+\frac{h_a+h_b}{h_a-2r+h_b})$ $\geq \frac{8}{3}(\frac{r_a}{2r_a-r+h_a}+\frac{r_b}{2r_b-r+h_b}+\frac{r_c}{2r_c-r+h_c})$ $\geq \frac{4}{3}(\cos^2{\frac{C}{2}}+\cos^2{\frac{A}{2}}+\cos^2{\frac{B}{2}})$ $\geq 4(\frac{\tan{\frac{B}{2}}}{\cot{\frac{C}{2}}+\tan{\frac{B}{2}}}+\frac{\tan{\frac{C}{2}}}{\tan{\frac{C}{2}}+\cot{\frac{A}{2}}}+\frac{\tan{\frac{A}{2}}}{\tan{\frac{A}{2}}+\cot{\frac{B}{2}}})$ $\geq \frac{4r}{h_b-2r+h_c}+\frac{4r}{h_a-2r+h_c}+\frac{4r}{h_a-2r+h_b}$ $ \geq \frac{a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)}{abc}$

Wlog, let $ a\ge\ b\ge c $. Then $ a(b+c-a)\le\b(c+a-b)\le\c(a+b-c) $ so by Rearrangement inequality, $ a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le\ ac(b+c-a)+ba(c+a-b)+bc(a+b-c) $ and $ a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le\ ab(b+c-a)+bc(c+a-b)+ac(a+b-c) $ so summing up the last two inequalities we get the desired inequality.

Wlog, let . Then so by Rearrangement inequality, , (1)and , (2)so summing up the last two inequalities we get the desired inequality.Password: Solution : From $(1)+(2), $ we have : $\sum a^{2}(b+c-a)\leq (ac+ab)(2p-2a)+(ab+bc)(2p-2b)+(bc+ac)(2p-2c)=2p\sum a(b+c)-2\sum ab(a+b)= 4p\sum ab-4p\sum ab+6abc=6abc , (*) $ In any triangle $ABC $ are valid the followings identities : $a+b+c=2p, abc=4Rrp, a^{2}+b^{2}+c^{2}=2p^{2}-2r^{2}-8Rr, a^{3}+b^{3}+c^{3}=2p(p^{2}-3r^{2}-6Rr) $. We have : (*) $\sum a^{2}(b+c-a)\leq 3abc\Leftrightarrow 2p\sum a^{2}-2\sum a^{3}\leq 3abc 2p(2p^{2}-2r^{2}-8Rr)-2p(2p(p^{2}-3r^{2}-6Rr)=-4pr^{2}-16Rrp+12pr^{2}+24Rrp\leq 12Rrp\Leftrightarrow 8pr^{2}+8Rrp-12Rrp\leq 0\Leftrightarrow 4pr(R-2r)\geq 0, $ Euler's

Here is my solution for Japanese Kids who (including me? ) don't know Schur, Rearrangement or Ravi's substitution. $3abc-a^2(b+c-a)-b^2(c+a-b)-c^2(a+b-c)$ $=a^3+b^3+c^3+3abc$ $-a^2(b+c)-b^2(c+a)-c^2(a+b)$ $=a^3+b^3+c^3-3abc+2abc+2abc+2abc-a^2(b+c)-b^2(c+a)-c^2(a+b)$ $=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ $-a(b^2-2bc+c^2)-b(c^2-2ca+a^2)-c(a^2-2ab+b^2)$ $=\frac 12(a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\}-a(b-c)^2-b(c-a)^2-c(a-b)^2$ $=\frac 12(b+c-a)(a-b)^2+\frac 12(c+a-b)(b-c)^2+\frac 12(a+b-c)(c-a)^2\geq 0.$ $ (\because a+b>c,\ b+c>a,\ c+a>b).$

See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=598883&hilit=IMO+1964

My Solution. By Schur's Inequality of degree 1, we get $-$ \begin{align*} &\sum_{\text{cyc}}a(a-b)(a-c)\ge0. \\\implies&(a^3-a^2b-a^2c+abc)+(b^3-b^2a-b^2c+abc)+(c^3-c^2a-c^2b+abc)\ge 0. \\\implies&(a^2b+a^2-a^3)+(b^2c+b^2a-b^3)+(c^2a+c^2b-c^3)\le 3abc. \\\implies&a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc. \end{align*}This completes the proof. $\square$ PS. We can apply Schur as $a,b,c$ are sides of a triangle, so they are positive real number. The inequality is also true for non-negative reals.

This is equivalent to $$\cos A+\cos B+\cos C\le\frac{3}{2}$$Which is the well-known result.

Another old one. Nice, very nice indeed. Of course, Ravi. We let $a=x+y,b=y+z,c=z+x.$ Thus the given inequality degenerates into \[ 2 \sum_{cyc} x(y+z)^2 \leq 3\sum_{cyc} xy(x+y) + 6xyz. \]The $xyz$ terms cancel neatly and so do a few other trivia, leaving us with \[ \sum_{cyc} xy(x+y) \geq 0.\]Which is clearly true as $x,y,z \geq 0.$

Beginer_vn86 wrote: This is equivalent to $$\cos A+\cos B+\cos C\le\frac{3}{2}$$Which is the well-known result. That's IMO Longlist 1970 task 10

Suppose: $c= \min\left \{ a,\,b,\,c \right \}$ Then: $$3\,abc- a^{2}\left ( b+ c-a \right )- b^{2}\left ( c+ a- b \right )- c^{2}\left ( a+ b- c \right )= \left ( a+ b- c \right )\left ( a- b \right )^{2}+ c\left ( a- c \right )\left ( b- c \right )\geqq 0$$

DPopov wrote: Suppose $a,b,c$ are the sides of a triangle. Prove that \[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc \] $$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc \iff (a+b-c)(a+c-b)(b+c-a)\leq abc$$$$a(b+c-a)+b(a+c-b)+c(a+b-c) >2abc \iff (a+b-c)(a+c-b)(b+c-a)>0$$here here

imagine $a \ge b \ge c$ then derivative of the function with respect to a is equal to $3bc-2ac-b^2-2a(b+c-a)$ which is always negative so we can put a=b then we shall have: $a^2c \ge c^2(2a-c)$ which is actually obvious!!!!!!!!

By Schur: $\sum_{cyc}{a(a-b)(a-c)} \ge 0$ $\iff a^3+b^3+c^3-a^2c-a^2b-b^2a-b^2c-c^2a-c^2b+3abc \ge 0$ $\iff a^2c+a^2b+b^2a+b^2c+c^2a+c^2b-a^3-b^3-c^3 \leq 3abc$ $\iff a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc$.$\blacksquare$

Let a=x+y , b=y+z , c=z+x. we have to prove: 2z(x+y)^2 + 2x(y+z)^2 + 2y(z+x)^2 ≤ 3(x+y)(y+z)(z+x) or in fact : 6xyz ≤ x^2(y+z) + y^2(z+x) + z^2(x+y) which can be proved with AM GM.

DPopov wrote: Suppose $a,b,c$ are the sides of a triangle. Prove that \[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc \] Comments: Whoa! It's a fun one. Proof. Subbing in $a=x+y, b=y+z, c=z+x \forall x,y,z > 0,$ the claim is: $$6xyz \leq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2,$$which is a direct consequence of AG-GM. Q.E.D

$a=x+y ; b=y+z ; c=x+z$ (because a,b,c are triangle's side) after We simplified : $6xyz\le \sum x(y^2+z^2)$ (1) We prove (1) by $AM \ge GM$ $Proved by : O.Y.SH.$

@above what do you mean by O.Y.SH?

O.Y.SH. = Ortikov Yusufjon Shamshodovich

Let $D,E,F$ be the intouch points of the triangle $ABC$ (with $D$ opposite $A$, $E$ opposite $B$), and let $x=AE=AF$, $y=BD=BF$, and $z=CD=CE$. Then $a=x+y$, $b=y+z$, and $c=z+x$. We need to show that: $$2z(x+y)^2+2x(y+z)^2+2y(z+x)^2\le3(x+y)(y+z)(z+x),$$which, after expansion, is: $$2\sum_{\text{sym}}x^2y+12xyz\le3\sum_{\text{sym}}x^2y+6xyz$$$$\Leftrightarrow\sum_{\text{sym}}x^2y\ge6xyz$$which is true by AM-GM.

The solution is even more obvious if we use Schur's Inequality.

Since $a,b,c$ are the sides of a triangle, there exist some nonnegative reals $x,y,z$ such that $a=y+z$, $b=z+x$, and $c=x+y$. Plugging in, the LHS becomes $$\sum_{\text{cyc}}(y+z)^2(2x)=\sum_{\text{cyc}}(2xy^2+2xz^2+4xyz)=2\sum_{\text{sym}}xy^2+12xyz,$$while the RHS becomes $$3(y+z)(z+x)(x+y)=3\left(\sum_{\text{sym}}xy^2+2xyz\right)=3\sum_{\text{sym}}xy^2+6xyz,$$so it suffices to prove that $$2\sum_{\text{sym}}xy^2+12xyz\leq3\sum_{\text{sym}}xy^2+6xyz\iff6xyz\leq\sum_{\text{sym}}xy^2.$$But by the AM-GM Inequality, $$\sum_{\text{sym}}xy^2\geq6\sqrt[6]{\prod_{\text{sym}}xy^2}=6\sqrt[6]{x^6y^6z^6}=6xyz,$$hence proven. Equality holds at $x=y=z\iff a=b=c$. $\blacksquare$

$\blacksquare$

Attachments:

https://artofproblemsolving.com/community/c6h598883p3556328 \[ a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc \]\[\iff abc \ge (a+b-c)(a+c-b)(b+c-a).\]

I have discussed this problem in my Ravi's transformation video on the inequalities playlist on my youtube channel "little fermat" . Here is the Link

Ravi sub, expand and muirhead.

By Using $\textbf{Schur's inequality}(n=1)$ we have: $a^3 + b^3 + c^3 + 3abc \geq \sum (a^2b)+ \sum (ab^2)$ $\Leftrightarrow \sum (a^2b) + \sum (ab^2) - \sum (a^3) \leq 3abc$ and it's equivalent to $shur(n=1) \blacksquare .$