zschess wrote: Let $ a, b, c $ be the side lengths of a triangle. Prove that $$ \displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 2(a^2 + b^2 + c^2) $$ Even $\displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 4(a^2 + b^2 + c^2)$ is true for all positives $a$, $b$ and $c$.

arqady wrote: zschess wrote: Let $ a, b, c $ be the side lengths of a triangle. Prove that $$ \displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 2(a^2 + b^2 + c^2) $$ Even $\displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 4(a^2 + b^2 + c^2)$ is true for all positives $a$, $b$ and $c$. We have \[\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b}- 4(a^2 + b^2 + c^2)=\sum{\frac{(a^2-2ab+bc)^2}{ab}}\ge{0}\]

zschess wrote: Let $ a, b, c $ be the side lengths of a triangle. Prove that $$ \displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 2(a^2 + b^2 + c^2) $$ $$ \displaystyle \frac{(a^2 + b^2)(a + c)}{b} \ge (a^2 + b^2 ) $$.Now sum up.

arqady wrote: zschess wrote: Let $ a, b, c $ be the side lengths of a triangle. Prove that $$ \displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 2(a^2 + b^2 + c^2) $$ Even $\displaystyle\sum_{cyc} \frac{(a^2 + b^2)(a + c)}{b} \ge 4(a^2 + b^2 + c^2)$ is true for all positives $a$, $b$ and $c$. This is a nice one. First we use Ravi substitution. We are left to show $ \displaystyle\sum_{cyc} \frac {[(y+z)^2+(x+z)^2][(y+z)+(x+y)]}{x+z} \ge 8(x^2+y^2+z^2+xy+yz+zx)$ This boils down to showing $ \displaystyle\sum_{cyc} \frac {y[y+z]^2}{z+x} \ge 2(x^2+y^2+z^2)$. Applying Cauchy to the above it suffice to show $(a+b)^2 \ge 4ab$ . Here we have assumed that $a=x^2+y^2+z^2$ $b=xy+yz+xz$.

hey we have (a+c)>b or, (a+c)/b>1 or, (a^2+b^2)(a+c)/b>a^2+b^2 similarly, (b^2+C^2)(b+a)/c>b^2 +c^2 and (c^2+a^2)(c+b)/a>c^2+a^2 so adding we have LHS>RHS i must be wrong somewhere because i didnt get the equality please help by pointing it out

AS a+c>b therefore (a^2 + b^2)(a+c)>b(a^2+b^2) and dividing by b we add all three cases and we are done