\begin{align} &2(ab+bc+ca)\le 5+ abc\nonumber\\ \Longleftrightarrow&2(a+b+c)\sum ab\le15+3abc\\ \Longleftrightarrow&2\sum(a^2b+ab^2)+6abc\le3abc+15\\ \Longleftrightarrow&2\sum(a^2b+ab^2)+3abc\le\frac59(a+b+c)^3\\ \Longleftrightarrow&\frac53(a^3+b^3+c^3)+abc\ge\sum(a^2b+ab^2)\\ \Longleftarrow&a^3+b^3+c^3+3abc\ge\sum(a^2b+ab^2) \end{align}(1) obtained by multiplying $3$ or $a+b+c$ to each term. (2) obtained by expanding. (3) obtained by cancelling terms and using $a+b+c=3$ again. (4) obtained by expanding. (5) obtained by the fact that $a^3+b^3+c^3\ge3abc$, and is Schur.

parmenides51 wrote: Let $a,b,c\in R^+$ with $a+b+c=3$. Prove that $$2(ab+bc+ca)\le 5+ abc$$(Real Matrik) pqr method: Let $p=a+b+c=3, q = ab + bc + ca, r = abc$. We need to prove that $2 q \le 5 + r$. Using degree three Schur $r\ge \frac{4pq - p^3}{9}$ and $q \le p^2/3 = 3$, we have $5 + r - 2q \ge 5 + \frac{4pq - p^3}{9} - 2q = 2 - \frac23 q \ge 0$.

Let $K = \{ (a,b,c) \in \mathbb{R}^3 \colon a,b,c \ge 0 , \, a+b+c=3 \}$ and $f \colon K \to \mathbb{R}$, $f(a,b,c) = 2(ab+bc+ca)-5-abc$. By compactness, $f$ has a global maximum $(x, y, z ) \in K$. We show that $x=y=z$, then $f(x,y,z)=f(1,1,1)=0$, as desired. Indeed, we can assume wlog that $x \le y \le z$. Then $x,y < 2$. Write $$f(x,y,z) = 2 yz + 2x(3-x)-5-xyz = (2-x) yz + 2x(3-x) -5.$$If $y \not= z$ then $yz < \left( \textstyle\frac{y+z}{2} \right) ^2$. Since $x < 2$ this implies $f(x, \textstyle\frac{y+z}{2} , \frac{y+z}{2} ) > f(x,y,z)$, a contradiction to the maximality. Hence, $y=z$ and analogously, by switching the roles of $x$ and $y$, we have $x=z$.

parmenides51 wrote: Let $a,b,c\in R^+$ with $a+b+c=3$. Prove that $$2(ab+bc+ca)\le 5+ abc$$(Real Matrik) WLOG, assume that $(b - 1)(c - 1)\ge 0$ (i.e. $b$ and $c$ are on the same side of $1$). We have \begin{align*} &5 + abc - 2(ab + bc + ca)\\ =\,& 5 + abc - 2(ab + bc + ca) + (a + b + c)(a + b + c - 3)\\ =\,& a^2 + b^2 + c^2 + abc - 3(a + b + c) + 5\\ =\,& \frac12(a + b + c- 3)^2 + \frac12(a - 1)^2 + a(b - 1)(c - 1) + \frac12 (b-c)^2\\ \ge\,& 0. \end{align*}

parmenides51 wrote: Let $a,b,c\in R^+$ with $a+b+c=3$. Prove that $$2(ab+bc+ca)\le 5+ abc$$(Real Matrik) https://artofproblemsolving.com/community/c6h539010p3100109

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