hello, for a) in the case of $ k=1$ we have $ {(a+b+c)(ab+bc+ca)-(a^3+b^3+c^3)=a^2b+3abc+ca^2+ab^2+b^2c+bc^2+c^2a-a^3-b^3-c^3=}$ $ 18xyz+5(y^2x+y^2z+yz^2+z^2x+x^2y+x^2z)>0$ for $ a=y+z,b=x+z,c=x+y$. Sonnhard.

plugging x+y,y+z,z+x into a,b,c it becomes following one $ 2(x + y + z)((k - 1)(x^2 + y^2 + z^2) + (3k + 1)(xy + yz + xz))\geq 3(x + y)(y + z)(z + x)$ if $ k<1$ we can choose $ x$ very large and $ y=z=0$. so we get a contradiction $ k$ cant be smaller than $ 1$.

a) \[ (a+b+c)(ab+bc+ca) >a^3 +b^3 +c^3\] \[ \iff a^2(b+c-a) +b^2(a+c-b) +c^2(a+b-c) +3abc >0\] I hate problems which are a) and b)

enndb0x wrote: a) \[ (a + b + c)(ab + bc + ca) > a^3 + b^3 + c^3\] \[ \iff a^2(b + c - a) + b^2(a + c - b) + c^2(a + b - c) + 3abc > 0\] I hate problems which are a) and b) Me too!

ridgers wrote: Let's consider the inequality $ a^3 + b^3 + c^3 < k(a + b + c)(ab + bc + ca)$ where $ a,b,c$ are the sides of a triangle and $ k$ a real number. a) Prove the inequality for $ k = 1$. b) Find the smallest value of $ k$ such that the inequality holds for all triangles. a) is obvious if you let $ a=x+y,b=y+z,c=z+x,x,y,z\ge 0$ and expand. For $ b$ let $ a=b=n,c=1$. Then we have: $ k > \frac{2n^3+1}{(2n+1)(n^2+2n)} = \frac{2n^3+1}{2n^3+5n^2+2n} = \frac{2+n^{-3}}{2+5n^{-1}+2n^{-2}} \to \frac{2}{2}=1$ when $ n \to \infty$ which gives $ k \ge 1$, and so the answer is $ k=1$.

My goodness Why do our leaders always recycle problems from previous olympiads? This one is a Bulgarian problem from 1998.

Hmmm, ok this is my solution Lets rewrite the inequality: (a+b+c)(ab+bc+ca)>a^3 + b^3 + c^3 having tha a,b,c are sides of triangle, we have a+b>=c so well have (a+b+c)(ab+bc+ca)>=(2c)(ab+c(a+b))=2abs+2c^3 now, lets prove that: 2abc+2c^3>a^3 + b^3 + c^3 <=> 2abc+c^3>a^3+b^3 Let c>=b>=a we'd have c^3>=b^3 so if 2abc>a^3 the inequality is proved so b>a, c>a,2a>a ======> 2abc>a^3 ....... and the other explonations,,,hmmmmmwhat dio u think, am i right???

Ledio9494 wrote: Hmmm, ok this is my solution Lets rewrite the inequality: (a+b+c)(ab+bc+ca)>a^3 + b^3 + c^3 having tha a,b,c are sides of triangle, we have a+b>=c so well have (a+b+c)(ab+bc+ca)>=(2c)(ab+c(a+b))=2abs+2c^3 now, lets prove that: 2abc+2c^3>a^3 + b^3 + c^3 <=> 2abc+c^3>a^3+b^3 Let c>=b>=a we'd have c^3>=b^3 so if 2abc>a^3 the inequality is proved so b>a, c>a,2a>a ======> 2abc>a^3 ....... and the other explonations,,,hmmmmmwhat dio u think, am i right??? In a triangle $ a+b>c$ not $ a+b\geq c$!

a) is trivial, for b) let put that $ a = x + y$ , $ b = y + z$ and $ c = z + x$ then $ x,y,z\geq 0$ we have then $ \sum a^3 = \sum(x + y)^3 = 2\sum x^3 + 3\sum xy(x + y)$ and we have $ (a + b + c)(ab + bc + ca) = 2(x + y + z)(\sum(x + y)(y + z))$ $ = 2(x + y + z)(3(xy + yz + zx) + (x^2 + y^2 + z^2))$ we have then $ k > \frac {a^3 + b^3 + c^3}{(a + b + c)(ab + bc + ca)} = \frac {2\sum x^3 + 3\sum xy(x + y)}{6(x + y + z)(xy + yz + zx) + (x + y + z)(x^2 + y^2 + z^2)}$ $ \Leftrightarrow k - 1 > - \frac {5\sum xy(x + y) + 18xyz}{18xyz + 8\sum xy(x + y) + 2\sum x^3}$ $ \Rightarrow k_{min} = 1$ It will be nicer, if the add a question c such that: c). find the greatest constant $ k$ for which the inequality given does not holds for all triangle

to endbox

peine wrote: a) is trivial, for b) let put that $ a = x + y$ , $ b = y + z$ and $ c = z + x$ then $ x,y,z\geq 0$ we have then $ \sum a^3 = \sum(x + y)^3 = 2\sum x^3 + 3\sum xy(x + y)$ and we have $ (a + b + c)(ab + bc + ca) = 2(x + y + z)(\sum(x + y)(y + z))$ $ = 2(x + y + z)(3(xy + yz + zx) + (x^2 + y^2 + z^2))$ we have then $ k > \frac {a^3 + b^3 + c^3}{(a + b + c)(ab + bc + ca)} = \frac {2\sum x^3 + 3\sum xy(x + y)}{6(x + y + z)(xy + yz + zx) + (x + y + z)(x^2 + y^2 + z^2)}$ $ \Leftrightarrow k - 1 > - \frac {5\sum xy(x + y) + 18xyz}{18xyz + 8\sum xy(x + y) + 2\sum x^3}$ $ \Rightarrow k_{min} = 1$ It will be nicer, if the add a question c such that: c). find the greatest constant $ k$ for which the inequality given does not holds for all triangle

to endbox

solution for $a)$ $a^3<a(ab+ac)=a^2(b+c)$ $b^3<b(ab+bc)$ and also for c and we get $3abc$ as rest in $RHS$, my solution for $b$ was wrong

My ideea for point b) after making the subsitutions. We have $2\displaystyle\sum\limits_{\text{cyc}}x^3+3\displaystyle\sum\limits_{\text{cyc}}(x^2y+xy^2)<2k\displaystyle\sum\limits_{\text{cyc}}x^3+\displaystyle\sum\limits_{\text{cyc}}16k(x^2y+xy^2)+36xyzk$, but we take $x,y,z$ positive arbitary reals. So we can take $y-0, z-0$ and $x=100$, of course $k\ge1$.("-"means it is very close(i don't know latex very well) Observe that we don't even care about the coefficients of $xyz$ and $x^2y,xy^2$,etc...

By $ s-R-r $ $ a^{3}+b^{3}+c^{3}=2s(s^{2}-6Rr-6r^{2}),ab+bc+ca=s^{2}+4Rr+r^{2} ,a+b+c=2s $ When $ k=1 $ The inequ be $ s^{2}-6Rr-6r^{2}<s^{2}+4Rr+r^{2} $ is obvious! Consider $ s^{2}-6Rr-6r^{2}<k(s^{2}+4Rr+r^{2}) ,(k<1) $ $ (1-k)s^{2}-(6+4k)Rr-(6+k)r^{2}<0 $ or $ s^{2}<\frac{6+4k}{1-k}Rr+\frac{6+k}{1-k}r^{2} $ We have $ 3\sqrt{3}r\leq s\leq \frac{3\sqrt{3}R}{2} $ and may have $ \frac{s}{R}\rightarrow 0 $, and when $ \frac{s}{R}\rightarrow 0 $,then $\frac{r}{R}\rightarrow 0 $,$ \frac{r}{s}\rightarrow 0 $ then $ \frac{\frac{6+4k}{1-k}Rr+\frac{6+k}{1-k}r^{2}}{Rs}\rightarrow 0 $ so $ k<1 $ is impossible! the smallest value of $ k $ is $ 1 $.

This problem is from Bulgarian competition from 7-8 years ago...

a) Upon the classical substitution $a=y+z, b=z+x, c=x+y$ the inequality becomes $5 \sum _{\text {sym}}x^2 y+18xyz>0$, which trivially holds. b) Suppose without loss of generality that $a \geq b \geq c$. For $c \rightarrow 0$ we have $a \rightarrow b$ and so $k>\frac {(a+b+c)(ab+bc+ca)} {a^3+b^3+c^3}\rightarrow \frac {ab(a+b)} {a^3+b^3} \rightarrow 1$, so that $k \geq 1$ and our desired minimal value for $k$ is $1$.

For the first part it is equialent to $a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc>a^3+b^3+c^3$ here Just use $\text{Triangle inequality }$ $a+b>c$ $b+c>a$ $c+a>b$

Mathias_DK wrote: ridgers wrote: Let's consider the inequality $ a^3 + b^3 + c^3 < k(a + b + c)(ab + bc + ca)$ where $ a,b,c$ are the sides of a triangle and $ k$ a real number. a) Prove the inequality for $ k = 1$. b) Find the smallest value of $ k$ such that the inequality holds for all triangles. a) is obvious if you let $ a=x+y,b=y+z,c=z+x,x,y,z\ge 0$ and expand. For $ b$ let $ a=b=n,c=1$. Then we have: $ k > \frac{2n^3+1}{(2n+1)(n^2+2n)} = \frac{2n^3+1}{2n^3+5n^2+2n} = \frac{2+n^{-3}}{2+5n^{-1}+2n^{-2}} \to \frac{2}{2}=1$ when $ n \to \infty$ which gives $ k \ge 1$, and so the answer is $ k=1$. hi Mathias_DK. is it true that k=1 ???? is it official solution