We could use $ uvw $ theorem.

Yes of course!

arqady wrote: Yes of course! in the exam we don't use uvw theorem .

alibez wrote: in the exam we don't use uvw theorem . Why?

i think we could use everything we wanted but we should proof things that wasn't in the lessons.

mmaht wrote: i think we could use everything we wanted but we should proof things that wasn't in the lessons. But a proof takes two lines and one figure: draw graphs of $f$ and $g$, where $f(x)=x^3-3ux^2+3v^2x$ and $g(x)=w^3$. We see that an extremal value of $w^3$ (if we want that the equation $f(x)=g(x)$ has tree positive roots) should be, when two numbers from $\{a,b,c\}$ are equal. We need to check also the case $w^3\rightarrow0^+$. All this very easy to check and very easy to write.

There is an elementary and short proof. There is no need for uvw. To prove $a+b+c+abc\le 3$, it suficces to note that not all $a,b,c$ can be greater or equal to 1. Therefore, wlog we can assume $c\le 1$ and that implies $abc\le ab\le ab+ac+bc$. Conclusion follows and equality holds when $a=3$, $b=c=0$ and all permutations. For inequality $a+b+c+abc\ge 2$, let's denote $a+b+c=p$, $ab+ac+bc=q$ and $abc=r$. If $p>2$, inequality is obviously true. Otherwise, if $p\le 2$, condition and am-gm imply: $3=p+q\le p+\frac{p^2}{3} \Longrightarrow p\ge \frac{3}{2}(-1+\sqrt{5})$ Now, Schur's inequality gives: $r\ge \frac{4pq-p^3}{9}$ , so it remains to prove: $p+\frac{4pq-p^3}{9} \ge 2$ But using $q=3-p$, inequality is equivalent to: $(2-p)(p^2+6p-9)\ge 0$ What is clearly true for $p\in \left[\frac{-3+3\sqrt{5}}{2}, 2 \right]$. Equality holds for $a=b=1$, $c=0$ and permutations. $\blacksquare$

for #1 with a+b+c+ab+bc+ca=3 we have \[3-(a+b+c+abc)=\sum\frac{c(a-b)^2}{9}+\frac{(a+b+c-3)^2}{3}+\frac{2(a+b+c)(ab+bc+ca)}{9}\] \[a+b+c+abc-2=\sum\frac{c(ab+a+b-1)^2}{8}+\sum\frac{c(a-1)^2(b-1)^2}{24}+\sum\frac{(a-1)^2(b-c)^2}{36}\]

szl6208 wrote: for #1 with a+b+c+ab+bc+ca=3 we have \[3-(a+b+c+abc)=\sum\frac{c(a-b)^2}{9}+\frac{(a+b+c-3)^2}{3}+\frac{2(a+b+c)(ab+bc+ca)}{9}\] \[a+b+c+abc-2=\sum\frac{c(ab+a+b-1)^2}{8}+\sum\frac{c(a-1)^2(b-1)^2}{24}+\sum\frac{(a-1)^2(b-c)^2}{36}\] Nice! You now call me! BQ

With Mixing I think we can solve it

szl6208 wrote: for #1 with a+b+c+ab+bc+ca=3 we have \[3-(a+b+c+abc)=\sum\frac{c(a-b)^2}{9}\]\[+\frac{(a+b+c-3)^2}{3}+\frac{2(a+b+c)(ab+bc+ca)}{9}\] \[a+b+c+abc-2=\sum\frac{c(ab+a+b-1)^2}{8}\]\[+\sum\frac{c(a-1)^2(b-1)^2}{24}+\sum\frac{(a-1)^2(b-c)^2}{36}\] $3-(a+b+c+abc)=\sum\frac{c(a-b)^2}{9}$ $+\frac{(a+b+c-3)^2}{3}+\frac{2(a+b+c)(ab+bc+ca)}{9}>0,$ $a+b+c+abc<3.$

$9(a+b+c)+9abc \le (a+b+c)(9+3-(a+b+c)) \le 27$ because $a+b+c \le 3$ If $a+b+c \ge 2$ then $a+b+c+abc \ge 2$ For $a+b+c \le 2$ By schur's inequality we have $(a+b+c)^3+9abc \ge 4(a+b+c)(3-(a+b+c))$ Then $9(a+b+c)+9abc-18 \ge 9(a+b+c)+4(a+b+c)(3-(a+b+c))-(a+b+c)^3-18\ge 0$ Because $3(\sqrt{2}-1)< \frac32(\sqrt{5}-1)\le a+b+c \le 2$

We wlog assume $c\le 1$ and that implies $a+b+c+abc\le a+b+c+ab<a+b+c+ab+bc+ca=3.$

《Mathematical communication》(China Wuhan) N0.9(2014) Q188: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc+abc= 7$, prove that \[ abc(a+b+c)\leq 3\]

with condition $a+b+c+ab+bc+ca+abc=7$,we have \[3-abc(a+b+c)=\sum\frac{bc(1+b+c)(a-1)^2}{6}+\frac{(abc-1)^2}{3}+\frac{(a+b+c-3)^2}{6}\]

The following inequalities are also true. 1. Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c+ab+ac+bc+3abc=9$. Prove that: \[ abc(a+b+c)\leq 3 \]

2. Let $a$, $b$ and $c$ be non-negative numbers such that $\frac{a+b+c}{23}+ab+ac+bc+abc=\frac{93}{23}$. Prove that: \[ abc(a+b+c)\leq 3 \] 3. Let $a$, $b$, $c$ and $k$ be non-negative numbers such that $a+b+c+k(ab+ac+bc)+abc=4+3k$. Prove that: \[ abc(a+b+c)\leq 3 \]

sqing wrote: 《Mathematical communication》(China Wuhan) N0.9(2014) Q188: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc+abc= 7$, prove that \[ abc(a+b+c)\leq 3\] $7=\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+ab+ac+bc+abc$ $\ge 7\sqrt[7]{(\frac{abc(a+b+c)}{3})^3}\implies abc(a+b+c)\leq 3$. The following inequalities are also true. For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc+abc= 7$, prove that \[ abc(a+b+c)^2\leq 9\].

sqing wrote: $\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+ab+ac+bc+abc$ $\ge 7\sqrt[7]{(\frac{abc(a+b+c)}{3})^3}$ Why?

arqady wrote: sqing wrote: $\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+ab+ac+bc+abc$ $\ge 7\sqrt[7]{(\frac{abc(a+b+c)}{3})^3}$ Why? But isn't it obvious from AM-GM Inequality?

YESMAths wrote: But isn't it obvious from AM-GM Inequality? How?

arqady wrote: YESMAths wrote: But isn't it obvious from AM-GM Inequality? How? Like this.

What are we doing with $ab+ac+bc$?

arqady wrote: sqing wrote: $\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+ab+ac+bc+abc$ $\ge 7\sqrt[7]{(\frac{abc(a+b+c)}{3})^3}$ Why? $\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3}+ab+ac+bc+abc$ $\ge 7\sqrt[7]{(\frac{(a+b+c)}{3})^3\cdot ab\cdot ac\cdot bc\cdot abc}=7\sqrt[7]{(\frac{abc(a+b+c)}{3})^3}.$

Thank you! I saw only $7=a+b+c+ab+ac+bc+abc\geq a+b+c+\sqrt{3abc(a+b+c)}+abc\geq$ $\geq7\sqrt[7]{\left(\frac{a+b+c}{3}\right)^3\left(\frac{abc(a+b+c)}{3}\right)^{\frac{3}{2}}abc}=7\left(\frac{a+b+c}{3}\right)^{\frac{9}{14}}(abc)^{\frac{5}{14}}$.

sqing wrote: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc+abc= 7$, prove that \[ abc(a+b+c)^2\leq 9\]. We can prove this inequality by $uvw$.

arqady wrote: The following inequalities are also true. 3. Let $a$, $b$, $c$ and $k$ be non-negative numbers such that $a+b+c+k(ab+ac+bc)+abc=4+3k$. Prove that:\[ abc(a+b+c)\leq 3 \] $4+3k=a+b+c+k(ab+ac+bc)+abc\ge (4+3k)\sqrt[4+3k]{(abc)^{2k+2}}$ $\implies abc\le 1$,hence $6+3k\ge a+b+c+k(ab+ac+bc)+3abc $ ${\ge (k+2)\sqrt[k+2]{3abc(a+b+c)(ab+ac+bc)^k}}$ ${\ge (k+2)\sqrt{3abc(a+b+c)}}$, $\implies abc(a+b+c)\leq 3$.

alibez wrote: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc= 3$, prove that \[2\leq a+b+c+abc\leq 3\]Proposed by Mohammad Ahmadi Let $c=min\{a,b,c\}$ , hence $c<1$ , $a+b+c+abc<a+b+c+ab<a+b+c+ab+ac+bc= 3$.

We could just finish by applying Lagrange Multipliers...Its pretty neat so i won't bother posting.

alibez wrote: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc= 3$, prove that \[2\leq a+b+c+abc\leq 3\] Proposed by Mohammad Ahmadi Could some one explain the Lagrange method solution please?

K.N wrote: alibez wrote: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc= 3$, prove that \[2\leq a+b+c+abc\leq 3\] Proposed by Mohammad Ahmadi Could some one explain the Lagrange method solution please? http://www.math.ksu.edu/~nagy/snippets/ineq-lagrange.pdf

sqing wrote: K.N wrote: alibez wrote: For any $a,b,c>0$ satisfying $a+b+c+ab+ac+bc= 3$, prove that \[2\leq a+b+c+abc\leq 3\] Proposed by Mohammad Ahmadi Could some one explain the Lagrange method solution please? http://www.math.ksu.edu/~nagy/snippets/ineq-lagrange.pdf Thanks I know what is Lagrange method I want to see the solution of this problem using this method !

$p,q,r:=a+b+c,ab+bc+ac,abc$.We have $p+q=3$ and it suffice to show $p+r\in [2,3]$.From third degree Schur we have :$$p^3+9r\geq 4pq$$switching $q=p-3$ this rearranges as $r\geq \frac{12p-4p^2-p^3}{9}$.If $p>2$ there's nothing to shows so for the sake of the first bound suppose $p\leq 2$ we need to show that $21p-4p^2-p^3-18\leq 0$ or $(2-p)(p^2+6p-9)\geq 0$.Obviously the expression in the first bracket is larger than zero.Note that $q\leq \frac{p^2}{3}$ and hence $p^2+3p-9>0$ and so as $p$ is positive we have $p\geq 3\left(\frac{\sqrt{5}-1}{2} \right) >3(\sqrt{2}-1)$ and hence $p^2+6p-9>0$ as well so the first bound is shown.Now $q^2\geq 3pr$ and so $p+\sqrt{3pr}\leq 3$ squaring we get $r\leq \frac{9-6p+p^2}{3p}$ and hence we need to show $9-15p+4p^2\leq 0$ or $(p-3)(4p-3)\leq 0$.The first bracket is less than or eqal zero while the second is larger as $p\geq 3\left(\frac{\sqrt{5}-1}{2}\right)$ and hence the bounds follow.

Just Use these: $q^2\geq 3pr$ $p^2\geq 3q$ $r\geq max\{0,\frac{4pq-p^3}{9}\}$

Let $a+b+c = p$ and $ab+bc+ca = q$ and $abc = r$ then we need to prove $2 \le p + r \le p + q$. part $1 : p+r \le p+q$ we need to prove $r \le q$. Assume that $q < r$ then $q^2 \ge 3pr > 3pq \implies q > 3p$ so $p^2 \ge 3q > 9p \implies p > 9$ which gives contradiction cause $p + q = 3$. part $2 : 2 \le p+r$ Note that $p^2 \ge 3q$ so $p + \frac{p^2}{3} - 3\ge 0$ and Note that $a^3 + b^3 + c^3 + 3abc \ge \sum{(a+b)ab}$ so $3r \ge \frac{4qp - p^3}{3}$ and putting $q = 3-p$ we need to prove $p + \frac{4(3-p)p - p^3}{9} \ge 2$ or $-p^3 - 4p^2 + 21p - 18 \ge 0$ or $(-p+2)(p^2 + 6p - 9) \ge 0$. Note that $p + \frac{p^2}{3} - 3\ge 0$ so for $p \le 2$ we're done. for $p > 2$ it's obvious that $p+r \ge 2$.