any ideas?

gobathegreat wrote: Let $a$, $b$, $c$, $d$ and $e$ be distinct positive real numbers such that $a^2+b^2+c^2+d^2+e^2=ab+ac+ad+ae+bc+bd+be+cd+ce+de$ $a)$ Prove that among these $5$ numbers there exists triplet such that they cannot be sides of a triangle Let us assume for the sake of contradiction that there are no triples among $a,b,c,d,e$ such that they cannot be sides of a triangle. Since, we can form a triangle with any three of $a,b,c,d, e$ as side lengths, the triangle inequality must hold for any three of $a,b,c,d,e$ . So, we have the following inequalities $$a+b >c \implies ac+bc > c^2 \quad, \quad \quad a+b > d \implies ad+bd > d^2 \quad ,\quad \quad a+b> e \implies ae+be > e^2 $$ $$a+c >b \implies ab+bc > b^2 \quad, \quad \quad a+c > d \implies ad+cd > d^2 \quad ,\quad \quad a+c> e \implies ae+ce > e^2 $$ $$a+d >b \implies ab+bd > b^2 \quad ,\quad \quad a+d > c \implies ac+cd > c^2\quad ,\quad \quad a+d> e \implies ae+de > e^2 $$ $$a+e >b \implies ab+be > b^2 \quad ,\quad \quad a+e > c \implies ac+ce > c^2 \quad ,\quad \quad a+e> d \implies ad+de > d^2 $$ $$b+c >a \implies ba+ca > a^2 \quad, \quad \quad b+c > d \implies bd+cd > d^2 \quad, \quad \quad b+c> e \implies be+ce > e^2 $$ $$b+d >a \implies ba+da > a^2 \quad ,\quad \quad b+d > c \implies bc+cd > c^2\quad, \quad \quad b+d> e \implies be+de > e^2 $$ $$b+e >a \implies ba+ea > a^2 \quad, \quad \quad b+e > c \implies bc+ec> c^2 \quad ,\quad \quad b+e> d \implies bd+ed > d^2 $$ $$c+d >a \implies ca+da > a^2 \quad ,\quad \quad c+d > b \implies cb+db> b^2 \quad ,\quad \quad c+d> e \implies ce+de > e^2 $$ $$c+e>a \implies ca+ea > a^2 \quad ,\quad \quad c+e > b \implies cb+eb> b^2 \quad, \quad \quad c+e> d \implies cd+ed > d^2 $$ $$d+e>a \implies da+ea > a^2 \quad ,\quad \quad d+e> b \implies db+eb> b^2 \quad ,\quad \quad d+e> c \implies dc+ec > c^2 $$ Adding all of the above inequalities, we obtain $$6(ab+ac+ad+ae+bc+bd+be+cd+ce+de) > 6(a^2+b^2+c^2+d^2+e^2)$$ $$ab+ac+ad+ae+bc+bd+be+cd+ce+de > a^2+b^2+c^2+d^2+e^2 $$ which is a contradiction since we are given that $$ab+ac+ad+ae+bc+bd+be+cd+ce+de = a^2+b^2+c^2+d^2+e^2$$ Therefore, we conclude that there must be a triplet among $a,b,c,d,e$ which cannot be sides of a triangle