Alternatively, completing squares, we get $ 5a^2 + 5b^2 + 5c^2 < 6ab + 6ac + 6bc \Longleftrightarrow (5a - 3b - 3c)^2 + 16(b - c)^2 < 16bc$. Now suppose $ a \ge b + c$. We have $ 16bc > (5b + 5c - 3b - 3c)^2 + 16(b - c)^2 = 4(5b^2 + 5c^2 - 6bc)$, which gives $ 10bc > 5b^2 + 5c^2$.

Albanian BMO TST 2010: 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. Find the smallest value of $ k$ such that the inequality holds for all triangles.

k = 10 to find it prove that, 2(ab + bc + ca) > a^2 + b^2 + c^2

fighter wrote: k = 10 to find it prove that, 2(ab + bc + ca) > a^2 + b^2 + c^2 err... I think jgnr's right. Triangle inequality should be invoked somehow.

The condition that $a$,$b$,$c$ are sides of triangle is equivalent to the condition $a=x+y$ , $b=y+z$ and $c=z+x$ for some positive real numbers $x$, $y$ and $z$. Therefore, we just need to find all $k$ such that , $$ k\sum_{cyc} (x+y)(y+z) \geq \sum_{cyc} 5(x+y)^2$$ $$\Leftrightarrow k\sum_{cyc} x^2 + 3k\sum_{cyc} xy \geq 10 \sum_{cyc} x^2 + 10 \sum_{cyc} xy $$ For $ 3 \geq k$, inequality becomes $$ 0 > (10-k)\sum_{cyc} x^2 + (10-3k)\sum_{cyc} xy$$ which is obviously false for all positive $x,y,z$. For $10 > k \geq 4$, inequality becomes $$ (3k-10)\sum_{cyc} xy \geq (10-k)\sum_{cyc} x^2$$ which is false for $x \to \infty$, $y=0$ and $z = 0$ as LHS $=0$ and RHS $\to \infty$ For $ k \geq 10$, the inequality ecomes $$ (k-10)\sum_{cyc} x^2 + (3k-10)\sum_{cyc} xy \geq 0$$ which is true for all positive real numbers $x$,$y$ and $z$. Therefore, for all postive integers $k \geq 10$ the inequality holds.

it is better to use the fact that \[ (ab + bc + ca) > =(a^2 + b^2 + c^2),\]\[ 5(ab + bc + ca) > 5(a^2 + b^2 + c^2),\]for the elimination of equality k>5 minimum value of k such that it is an integer and is greater than 5 is 6 k>6 such that k is an integer and \[ k(ab + bc + ca) > 5(a^2 + b^2 + c^2),\]

OG! For $k\geq7$ $a=0.25, b=0.5, c=1$ satisfies. So, no solution For $k \leq5$, the inequality $ k(ab + bc + ca) > 5(a^2 + b^2 + c^2)$ can be translated to $ (k-5)(ab + bc + ca) > 5(a^2 + b^2 + c^2- ab- bc - ca)$ Which is, $0\geq (k-5)(ab + bc + ca) > 5(a^2 + b^2 + c^2-ab + bc + ca)= 5( \frac{1}{2}(a-b)^2+\frac{1}{2}(b-c)^2+ \frac{1}{2}(c-a)^2)$ A contradiction! For $k=6$ We claim $k=6$ satisfies. for $k=6$, the inequality is $\frac{6}{5} > \frac{(a^2 + b^2 + c^2)}{a(b+c)+ bc}$, WLOG $a \geq b \geq c$. It is sufficient to show $a<b+c$. So FTSOC, assume $a \geq b+c$. then $\frac{(a^2 + b^2 + c^2)}{a(b+c)+ bc} \geq \frac{((b+c)^2 + b^2 + c^2)}{(b+c)^2+ bc} = 1+ \frac{b^2 + c^2}{b^2 + c^2+2bc}= 2- \frac{2bc}{b^2 + c^2+2bc} \geq 2- 0.5 =1.5> \frac{6}{5}$, contradicting the inequality, Hence $k=6$ satisfies and is the only positive integral solution. Remark: $k \in (5,7.5]$ gives only and all possible real values of $k$

Quantum_fluctuations wrote: The condition that $a$,$b$,$c$ are sides of triangle is equivalent to the condition $a=x+y$ , $b=y+z$ and $c=z+x$ for some positive real numbers $x$, $y$ and $z$. Therefore, we just need to find all $k$ such that , $$ k\sum_{cyc} (x+y)(y+z) \geq \sum_{cyc} 5(x+y)^2$$ $$\Leftrightarrow k\sum_{cyc} x^2 + 3k\sum_{cyc} xy \geq 10 \sum_{cyc} x^2 + 10 \sum_{cyc} xy $$ For $ 3 \geq k$, inequality becomes $$ 0 > (10-k)\sum_{cyc} x^2 + (10-3k)\sum_{cyc} xy$$ which is obviously false for all positive $x,y,z$. For $10 > k \geq 4$, inequality becomes $$ (3k-10)\sum_{cyc} xy \geq (10-k)\sum_{cyc} x^2$$ which is false for $x \to \infty$, $y=0$ and $z = 0$ as LHS $=0$ and RHS $\to \infty$ For $ k \geq 10$, the inequality ecomes $$ (k-10)\sum_{cyc} x^2 + (3k-10)\sum_{cyc} xy \geq 0$$ which is true for all positive real numbers $x$,$y$ and $z$. Therefore, for all postive integers $k \geq 10$ the inequality holds. Sir, the question doesn't ask for whether a,b,c are sides of triangles implies the inequality, it instead asks the opposite i.e. If the inequality holds, then for what $k$ a, b ,c are necessarily sides of a triangle.

Progress: $k \le 5$ all work because no such $a, b, c$ exist; $k \le 6$ by putting $(a, b, c) = (1, 1, 2)$. We need to check if $k = 6$ works or not.

@Above k =6 works SEE #9 post

sansgankrsngupta wrote: @Above k =6 works SEE #9 post I was planning to edit *after* I solved, thank you very much.