We will prove that the minimum is $3$ Indeed, note that: $A=\sum \frac{2 - a^3}{a}=\sum \frac{2}{a}-\sum a^2$ Using the initial condition: $\sum a^2=9-2(ab+bc+ca)$ $\sum\frac{2}{a}=\frac{2(ab+bc+ca)}{abc}$ $(ab+bc+ca)^2\geq 3abc(a+b+c)=9abc$ It suffices to prove: $\frac{18}{q}+2q\geq 6$, which is obvious by AM-GM. Equality holds iff $a=b=c=1$ Q.E.D

we have $\sum\frac{2 - a^3}{a}=\sum \frac{2}{a}-\sum a^2=\sum \frac{2}{a} +2\sum ab -9$ now by AM-GM we have $\sum \frac{2}{a}+2\sum ab \ge 12$ so that $A=\sum \frac{2}{a} +2\sum ab -9\ge 3$ where equality occurs iff $a=b=c=1$ so we are done

My solution So by basic transformations we get $A=2B-9$ So it is equivalent to find the min of $B$ where $B=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+ab+bc+ca$ By AM-GM $B^2\ge 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(ab+bc+ca)=4(2(a+b+c)+\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}) \ge12(a+b+c)=36$ So $B\ge6$ And $A=2B-9\ge12-9=3$ Equality holds for $a=b=c=1$

$A=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\{(a+b+c)^2-2(ab+bc+ca)\}$ $=2(ab+bc+ca)\left(\frac{1}{abc}+1\right)-9$ $\therefore (A+9)^2=4(ab+bc+ca)^2\left(\frac{1}{abc}+1\right)^2$ $\geq 4\cdot 3abc(ab+bc+ca)\left(\frac{1}{abc}+1\right)^2=36abc\left(\frac{1}{abc}+1\right)^2$ $\geq 36abc\cdot \frac{4}{abc}=12^2$, since $A+9>0$, we have $A+9\geq 12 $, yielding $A\geq 3.$ Equality holds when $a=b=c=1.$, therefore the minimum value of $A$ is 3.

Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Prove that$$\dfrac{2-a^5}{a^2}+\dfrac{2-b^5}{b^2}+\dfrac{2-c^5}{c^2}\geq3.$$

Who is deleting my posts on this topic? Is pointing out to wrong solutions getting banned on this forum? Why are you deleting "solutions" that come out to be incorrect. Learning from mistakes is also important. This is forum and not a magazine where solutions are listed below problem statement.

sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Prove that$$\dfrac{2-a^5}{a^2}+\dfrac{2-b^5}{b^2}+\dfrac{2-c^5}{c^2}\geq3.$$ Let $f(x)=\frac{2}{x^2}-x^3$. Hence, $f''(x)=\frac{12}{x^4}-6x$, which gives that $f$ is a convex function on $\left(0,\sqrt[5]{2}\right]$. Let $\{b,c\}\subset\left(0,\sqrt[5]{2}\right]$. Hence, by Jensen $f(b)+f(c)\geq2f\left(\frac{b+c}{2}\right)=2f\left(\frac{3-a}{2}\right)$. Thus, it remains to prove that $2\left(\frac{2}{\left(\frac{3-a}{2}\right)^2}-\left(\frac{3-a}{2}\right)^3\right)+\frac{2}{a^2}-a^3\geq3$, which is $(a-1)^2(24+32a-53a^2+21a^3+a^4-a^5)\geq0$, which is true even for all $a\in(0,3)$. Let $0<a\leq\sqrt[5]2\leq b\leq c$. Hence, by Karamata $\sum_{cyc}f(a)\geq f(a)+f\left(\sqrt[5]2\right)+f\left(b+c-\sqrt[5]2\right)$. Thus, it remains to prove that $\frac{2}{a^2}-a^3+\frac{2}{\left(3-\sqrt[5]2-a\right)^2}-\left(3-\sqrt[5]2-a\right)^3\geq3$, which is true even for all $a\in(0,1.5)$.

We can use also $uvw$: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Hence, our inequality is equivalent to $f(w^3)\geq0$,where $f$ is a decreasing function. Hence, $f$ gets a minimal value for maximal value of $w^3$, which happens for equality of two variables and the rest is easy.

Does anyone have idea about how many students could solve it on contest? It seems that 2nd problem was harder than the 3rd one, doesn' it?

About 10 % of all participants seemed to have been able to solve the problem.

My solution : $2/a+2/b+2/c -a^2-b^2-c^2\ge 3 $ we need to prove which is equvalent to $2/a+2/b+2/c+2ab+2bc+2ac\ge12$ $1/a+1/b+1/c+ab+ac+bc=(ab+ac+bc)/abc+ab+bc+ac\ge 3(a+b+c)/(ab+bc+ac)+ab +bc+ac $$(ab+bc+ac)^2\ge \sum3^2bc$ and we are basically done.

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+bc+ca+ab=\frac{bc+ca+ab}{abc}+bc+ca+ab$ $\geq\frac{3(a+b+c)}{bc+ca+ab}+bc+ca+ab=\frac{9}{bc+ca+ab}+bc+ca+ab\geq6,$ $\dfrac{2-a^3}a+\dfrac{2-b^3}b+\dfrac{2-c^3}c=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+bc+ca+ab\right)-9\geq3.$ http://www.artofproblemsolving.com/community/c6h1106910p5022568: Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Prove that$$\dfrac{1-a^3}a+\dfrac{1-b^3}b+\dfrac{1-c^3}c\geq -0.0724...$$ Stronger than Jbmo 2015

The minimum is $3$. Let $a=\frac{3x}{x+y+z},b=\frac{3y}{x+y+z},c=\frac{3z}{x+y+z}$.Then we have to prove that $2(x+y+z)^3(xy+yz+zx)\ge 9xyz(x+y+z)^2+27xyz(x^2+y^2+z^2)$.Let $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$.So we have to prove that $v^2(u^3+w^3)\ge 2w^3u^2$. It's easy to prove that $v^4\ge uw^3$,so $v^2(u^3+w^3)\ge \sqrt{uw^3}\cdot 2\sqrt{u^3w^3}=2w^3u^2$,which ends our proof.

Could someone please explain to me how in "aditya21" post above they got $\sum\frac{2}{a} + 2\sum ab \ge 12$ using AM-GM. Could you also please explain what I am doing wrong: $\sum\frac{2}{a} + 2\sum ab = \sum\frac{1}{a} + \sum\frac{1}{a} + \sum 2ab \ge 9 \times \sqrt[9]{\frac{1}{a.a.b.b.c.c} \times 2ab.2bc.2ca} \ge 9\times \sqrt[9]{8} \ge 9\sqrt[3]{2} \ne 12$

We have $\sum\frac{2}{a} + \sum 2ab \ge 12\sqrt[6]{abc}$ with AM-GM and equality when $a=b=c$. Or $a+b+c=3$ so $a=b=c=1$ and $\sum\frac{2}{a} + \sum 2ab \ge 12\sqrt[6]{abc}=12\sqrt[6]{1\times 1\times 1}=12$.

rgreenanswers wrote: Could someone please explain to me how in "aditya21" post above they got $\sum\frac{2}{a} + 2\sum ab \ge 12$ using AM-GM. Could you also please explain what I am doing wrong: $\sum\frac{2}{a} + 2\sum ab = \sum\frac{1}{a} + \sum\frac{1}{a} + \sum 2ab \ge 9 \times \sqrt[9]{\frac{1}{a.a.b.b.c.c} \times 2ab.2bc.2ca} \ge 9\times \sqrt[9]{8} \ge 9\sqrt[3]{2} \ne 12$ Your AM-GM's equality case is $\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=2ab=2bc=2ac$ which is impossible since we know $a+b+c=3$. This is where you're doing wrong

Ok that makes sense. Thanks to both of you!

It is sufficient to prove that $(ab+bc+ca)(abc+1)\geq6abc$, or $a^2b^2c+a^2bc^2+ab^2c^2+ab+bc+ca\geq6abc$. But by AM-GM, $a^2b^2c+ab\geq2\sqrt{a^3b^3c}=2ab\sqrt{abc}$. Thus, $LHS=\sum ab+\sum a^2b^2c\geq2\sqrt{abc}\sum ab \geq2\sqrt{abc}\sqrt{3abc(a+b+c)}=6abc$, $Q.E.D.$

Rearrange the expression as $2\sum\dfrac{1}{a} - \sum a^2 = 2\sum\dfrac{1}{a} - \left(\sum a\right)^2 + 2\sum ab = \left(2\sum ab\right)\left(\dfrac{1}{abc} + 1\right) - 9$. Note that $\dfrac{1}{abc} \ge 1$ by AM-GM, so our rearranged expression is greater than or equal to $4\left(\sum ab\right) - 9$. Applying AM-GM again yields $4\left(\sum ab\right) - 9 \ge 3$, and equality holds when $a=b=c=1$.

Can someone explain how to prove this? Quote: $\sum \frac{2}{a}+2\sum ab \ge 12$ by AM-GM Looking at the posts above, I do not see a valid proof of this, I may be mistaken.

neverlose wrote: Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Find the minimum value of the expression \[A=\dfrac{2-a^3}a+\dfrac{2-b^3}b+\dfrac{2-c^3}c.\] a,b,c>0,a+b+c=3 (9-8a^3)/a+(9-8b^3)/b+(9-8b^3)/b>=3

AM-GM is actually $\sum \frac{2}{a}+2\sum ab \ge 12 (abc)^{1/6} $ which is not enough to show the LHS is greater then or equal to one...?

Let's consider that $f(x)=\frac{2-x^3}{x}$. It's obvious that $f''(x)>0$ Hence we can attain $f(\frac{a+b+c}{3})\leq\frac{f(a)+f(b)+f(c)}{3}$ from Jensen Inequality $\Longrightarrow f(1)=1\leq\frac{f(a)+f(b)+f(c)}{3}\Longrightarrow 3\leq f(a)+f(b)+f(c)$

CrazyMathlete wrote: Let's consider that $f(x)=\frac{2-x^3}{x}$. It's obvious that $f''(x)>0$ Are you sure?

The answer is $\boxed{3}$, attained at $a=b=c=1$. First we take care of the case where one of $a,b,c$ is at least 2; WLOG let $c \geq 2$. Then $a,b \leq 1$ and since the $f$ is concave up on $(0,1]$ we have $f(a)+f(b) \geq 2 f( \frac{3-c}{2} )$; thus $f(a)+f(b)+f(c) \geq 2 f( \frac{3-c}{2} )+f(c)$ which, after some calculation, ends up greater than 3 for all $c \geq 2$. Now we show if all of $a, b, c$ are in the range $(0,2]$ then the quantity in the problem is at least $3$. I claim that $\frac{2-x^3}{x} \geq -4x+5$ for all $x \in (0,2]$. Indeed, after expansion and rearrangement it becomes $(x-1)^2(x-2) \leq 0$, which is obvious for $x \in (0,2]$ (equality occurs at $x=1$). Thus we have $$\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c} \geq -4(a+b+c)+15 = 3$$, as desired. We see equality occurs iff $a, b, c$ are all $1$.

silouan wrote: CrazyMathlete wrote: Let's consider that $f(x)=\frac{2-x^3}{x}$. It's obvious that $f''(x)>0$ Are you sure? Yes, you can check it. Btw,i did not see ur answer, sorry

I think this one was proposed by Greece!

neverlose wrote: Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Find the minimum value of the expression \[A=\dfrac{2-a^3}a+\dfrac{2-b^3}b+\dfrac{2-c^3}c.\]

$\blacksquare$

neverlose wrote: Let $a,b,c$ be positive real numbers such that $a+b+c = 3$. Find the minimum value of the expression \[A=\dfrac{2-a^3}a+\dfrac{2-b^3}b+\dfrac{2-c^3}c.\] let f（x）=（2-x^3）/x

Let $a,b,c$ be positive real numbers such that $a^2+2bc=3$. Prove that$$\dfrac{2-a^3}{a}+\dfrac{2-b^3}{b}+\dfrac{2-c^3}{c}\geq3.$$$$\dfrac{2-a^5}{a^2}+\dfrac{2-b^5}{b^2}+\dfrac{2-c^5}{c^2}\geq3.$$

Here is a quite painful solution using the method of linearization (tangent line trick). Perhaps some of the details can be simplified, but I doubt the simplification could be substantial. Let us firstly look for constants $m$ and $n$ such that $\frac{2-t^3}{t} \geq mt+n$ for $t\in (0,3)$ with equality at $t=1$. Equivalently, $t^3 + mt^2 + nt - 2 \leq 0$. In order for $1$ to be a double root, Horner's method leads to the equations $m+n = 1$ and $3 + 2m + n = 0$, whence $n=5$, $m=-4$ and the remaining multiplier is $t-2$, i.e. we have the equivalent $(t-2)(t-1)^2 \leq 0$, which unfortunately holds only for $t\in (0,2]$. In other words, if $a,b,c \in (0,2]$, then the expression is at least $3\cdot 5 - 4(a+b+c) = 3$ and we are done. From now on we will use that $\frac{2-t^3}{t}$ decreases for $t > 0$ since the numerator decreases while the positive denominator increases. Now suppose without loss of generality $c \in (2,3)$ and $a+b \in (0,1)$. Then $\frac{2-c^3}{c} > -\frac{25}{3}$ due to $(c-3)(3c^2+9c+2) < 0$. If it turns out that each of $\frac{2-a^3}{a}$ and $\frac{2-b^3}{b}$ exceeds $\frac{17}{3}$, then we will be done. Trial and error leads to $\frac{2-t^3}{t} \geq \frac{53}{9}$ for $t\in (0,\frac{1}{3}]$ (equivalent to $(3t-1)(3t^2+t+18) \leq 0$). So we are done if $a,b \leq \frac{1}{3}$. Let $b > \frac{1}{3}$, so $c \in (2, \frac{8}{3})$. Then $\frac{2-c^3}{c} > -\frac{229}{36}$ because of $(3t-8)(12t^2+32t+9) < 0$ and we shall be done if $\frac{2-a^3}{a} + \frac{2-b^3}{b} > \frac{337}{36}$. But $\frac{2-t^3}{t} \geq \frac{121}{25} > \frac{337}{72}$ for $t\in (0,\frac{2}{5}]$ due to $(5t-2)(5t^2+2t+25) \leq 0$. So for $a,b \leq \frac{2}{5}$ we are done. Let $b>\frac{2}{5}$. Then $c<\frac{13}{5}$ and $\frac{2-c^3}{c} > -6$, also $\frac{2-b^3}{b} > 1$ for $b<1$, so we are done if $\frac{2-a^3}{a} > 8$, i.e. for $a<\frac{6}{25}$. So we may assume $a\in (\frac{6}{25},\frac{1}{2}), b\in (\frac{2}{5}, \frac{19}{25})$, $c\in (2,\frac{59}{25})$. Then $\frac{2-c^3}{c} > -4$,$75$ while the other two exceed in total $3$,$875$ for $a,b < \frac{12}{25}$. So we are up to $a\in (\frac{6}{25},\frac{1}{2})$, $b \in (\frac{12}{25}, \frac{19}{25})$, $c \in (2, \frac{57}{25})$. Then $\frac{2-c^3}{c} > -4$,$35$ and we would like $\frac{2-a^3}{a} + \frac{2-b^3}{b} > 7$,$35$, which is true for $a < \frac{9}{25}$ and $b<\frac{19}{25}$. So now $a\in (\frac{9}{25}, \frac{1}{2})$, $b\in (\frac{12}{25}, \frac{16}{25})$, $c \in (2, \frac{54}{25})$. Respectively, $\frac{2-c^3}{c} > -3$,$74$ and we want $\frac{2-a^3}{a} + \frac{2-b^3}{b} > 6$,$74$, which holds for $a \leq \frac{47}{100}$, $b<\frac{16}{25}$. We have reached $a\in (\frac{47}{100}, \frac{1}{2}), b\in (\frac{12}{25}, \frac{53}{100})$, $c\in (2, \frac{41}{20})$. Here $\frac{2-a^3}{a} > \frac{15}{4}$, $\frac{2-b^3}{b} > 3$ and $\frac{2-c^3}{c} > -3$,$25$, so the main sum exceeds $3$, wohooo!

$LHS=2 \sum \frac{1}{a}-\sum a^2=2\sum \frac{1}{a}-((a+b+c)^2-2(ab+bc+ca)) =2\sum \frac{1}{a}+2\sum ab -9=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+ab+bc+ca)-9$ Let's prove that $\sum \frac{1}{a}+\sum ab \geq 6$ $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+(ab+bc+ca) \geq 6$ $\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+(ab+bc+ca)}{2} \geq 3$ By AM-GM $LHS \geq \sqrt[3]{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(ab+bc+ca)} \geq 3$ $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(ab+bc+ca) \geq 9$ $2(a+b+c)+\sum \frac{ab}{c} \geq 9$ $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \geq 3$ The last inequality is true because $LHS=\sum(\frac{ab}{2c}+\frac{bc}{2a}) \geq \sum 2 \sqrt{\frac{ab^2c}{4ac}}=\sum b=3$

I have A>=-9

i still dont understand the concept of chebyshev but can we apply it on the 2ab+2bc+2ca