Return to Revisions

This proof is similar to River Li's, but it requires no machine calculation. With the notation $$Q:=\frac{\cos^2\alpha+\cos^2\beta+\cos^2\gamma}{3} \qquad\text{and}\qquad R:=\cos\alpha\cos\beta\cos\gamma,$$ the polynomial in the original post equals $$p(t):=20t^3-30t^2+12(1-Q)t+(3Q-2R-1).$$ In particular, $$p(1/2+v)=20v^3-3(1+4Q)v-(3Q+2R).$$ We claim that this cubic polynomial has a nonnegative discriminant, that is, $$5(3Q+2R)^2\leq(1+4Q)^3.$$ To prove this inequality, we observe first that $$5(3Q+2R)^2=5(9Q^2+12QR+4R^2)\leq 5(15Q^2+10R^2)=25(3Q^2+2R^2).$$ Using also that $R^2\leq Q^3$ holds by the AM-GM inequality, it suffices to show that $$25(3Q^2+2Q^3)\leq (1+4Q)^3.$$ However, this one is clear, because $$(1+4Q)^3-25(3Q^2+2Q^3)=(Q-1)^2(1+14Q)\geq 0.$$ We proved that the cubic polynomial $p(1/2+v)$ has a nonnegative discriminant, so it has three real roots (counted with multiplicity). So the original polynomial $p(t)$ has three real roots (counted with multiplicity), and the same is true of $$25\cdot p(1/5+s)=500s^3-450s^2+60(1-5Q)s+(9+15Q-50R).$$ It remains to show that this last polynomial has a negative root. If this is not the case, then the polynomial has three nonnegative roots, hence in particular $$1-5Q>0>9+15Q-50R.$$ However, this contradicts the already observed inequality $R^2<Q^3$. Indeed, $R<Q^{3/2}\leq Q$, whence $$9<50R-15Q\leq 35Q<7,$$ which is a contradiction. The proof is complete.