1.Area of triangle by Heron's formula $\triangle=\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$ Therefore,we must have $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0$ $\left((b+c)^2-a^2\right)\left(-(b-c)^2+a^2\right)>0$ $a^2(b+c)^2+a^2(b-c)^2-a^4-(b+c)^2(c-b)^2>0$ $2(a^2b^2+b^2c^2+c^2a^2)>a^4+b^4+c^4$ $a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)>2(a^4+b^4+c^4)$ $ (a^2+b^2+c^2)^2>2(a^4+b^4+c^4) $ Alternatively Working backwards from given condition $ (a^2+b^2+c^2)^2>2(a^4+b^4+c^4) $ We get $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0$ Now if one of $(-a+b+c),(a-b+c),(a+b-c)$ is positive and the other two negative i.e.$c+a<b$ and $a+b<c$.On adding the two inequalities,we obtain $2a<0$,which is a contradiction. One of $(-a+b+c),(a-b+c),(a+b-c)$ cannot be negative as then the given condition $ (a^2+b^2+c^2)^2>2(a^4+b^4+c^4) $ would not happen. All three of $(-a+b+c),(a-b+c),(a+b-c)$ cannot be negative for the same reason. So,only possibility left is that all three of $(-a+b+c),(a-b+c),(a+b-c)$ must be positive,which satisfies the triangle inequality.

solution second part: let $a_1=max( a_i)$ and $a_1\ge a_2+a_3$ then $a_4^2+...+a_n^2=x$ and $a_4^4+...+a_n^4 \ge \frac{1}{n-3} x^2$. We now have $(a_1^2+a_2^2+a_3^2)^2 \le 2(a_1^4+b_1^4+c_1^4)$ and $ 2x(a_1^2+a_2^2+a_3^2) \le (n-3)(a_1^4+b_1^4+c_1^4) + \frac{2}{n-3} x^2$ $x^2=\frac{n-3}{n-3} x^2$ The sum gives $ (a_1^2+a_2^2+\dots+a_n^2)^2>(n-1)(a_1^4+a_2^4+\dots+a_n^4) $.

Iran 1996:Let $a, b, c$ be real numbers. Prove that there exists a triangle with side lengths $a, b, c$ if and only if \[2(a^4 + b^4 + c^4) < (a^2 + b^2 + c^2)^2.\]

created by murray klamkin If a,b,c three positive numbers satisfy the ralation:\[ a^2+b^2+c^2> \sqrt {2(a^4+b^4+c^4)} \]then a,b,c are the sides of a triangle!!

Tournament of Towns, Spring 1987: We are given three non-negative numbers $a ,b$ and $c$ about which it is known that $a^4 + b^4 + c^4 \le 2(a^2b^2 + b^2c^2 + c^2a^2)$ (a) Prove that each of $A, B$ and $C$ is not greater than the sum of the others. (b) Prove that$$a^2 + b^2 +c^2 \le 2(ab+bc+ ca)$$(c) Does the original inequality follow from the one in (b)?

SCP wrote: solution second part: let $a_1=max( a_i)$ and $a_1\ge a_2+a_3$ then $a_4^2+...+a_n^2=x$ and $a_4^4+...+a_n^4 \ge \frac{1}{n-3} x^2$. We now have $(a_1^2+a_2^2+a_3^2)^2 \le 2(a_1^4+b_1^4+c_1^4)$ and $ 2x(a_1^2+a_2^2+a_3^2) \le (n-3)(a_1^4+b_1^4+c_1^4) + \frac{2}{n-3} x^2$ $x^2=\frac{n-3}{n-3} x^2$ The sum gives $ (a_1^2+a_2^2+\dots+a_n^2)^2>(n-1)(a_1^4+a_2^4+\dots+a_n^4) $. I can’t get what you worked for the inequality.

Kinda unexpected induction ... Before we start we can let $a_i \geq a_j$ for every $i \leq j$. Observe that the objective is equivalent to proving $a_1 < a_{n-1}+a_n$, as if we can form a triangle with its "short edges" be the extreme smalls and the "longest edge" to be the extreme high we are set for life. For part (a), we consider a function on $a_1^2$: $f(x) = x^2 - 2x (a_2^2+a_3^2)+(a_2^4+a_3^4-2a_2^2a_3^2)$. This function is convex upwards, and so it has its lowest point at $x = a_2^2+a_3^2$. Given that $f(a_1^2)<0$, supposing that $a_1 \geq a_2+a_3$ we get $a_1^2 \geq a_2^2+a_3^2+2a_2a_3 > a_2^2+a_3^2$. So, we get \[ 0 > f(a_1^2) \geq f((a_2+a_3)^2) \]This is equivalent to $0 < 0$. Contradiction. For part (b) ($n \geq 4$), we isolate $a_1, a_{n-1}$ and $a_n$ and we consider $i \ne \{1,n-1,n\}$. Now we do the same thing with constructing a quadratic in $a_i^2$: Let \[f(x) = (n-2)x^2 - 2x (\sum_{j \ne i}aj^2 + (n-1) \sum_{j \ne i} a_j^4 - 2 \sum_{i \ne j \ne k \ne i} a_j^2a_k^2. \]This value attains its minimum at $a_i^2 = \sum_{j\ne i} \dfrac{a_j^2}{n-2}$, so we get \[ 0 > f(a_i^2) \geq f(\sum_{j\ne i} \dfrac{a_j^2}{n-2})\]From this fact (i.e. $f(some\, \, sum) < 0$), we attempt to compute $LHS - RHS$ when $x_i^2$ is fixed as above. Reverting $f(x)$ back to the form in the original inequality yields \begin{align*} LHS = (\dfrac{n-1}{n-2} \sum_{j\ne i} a_j^2)^2 &> (n-1) \left(\dfrac{(\sum_{j\ne i} a_j^2)^2}{(n-2)^2}+ \sum_{j \ne i} a_j^4 \right) = RHS \\ \dfrac{n-2}{(n-2)^2} (\sum_{j\ne i} a_j^2)^2 &> \sum_{j \ne i}a_j^4 \\ \sum_{j\ne i} (a_j^2)^2 &> (n-2) \sum_{j \ne i}a_j^4 \end{align*}After dividing by $n-1$ in both sides (from the first row of equation to the second row of equation). Therefore, by induction on $n-1$ variables $\{a_1,a_2, \ldots, a_n\}$ without $a_i$ we get that $a_1 < a_{n-1}+a_n$.