Show that $b^2+c^2-a^2\leq bc$.

Question

Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.

I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.

Answer 1

Both conditions and the hypothesis are homogenous, so we can scale the variables until $a=1$. Then the problem is equivalent to the implication: $$b<\sqrt{c}\land c<\frac{2b}{1+b}\Rightarrow b^2+c^2\le1+bc.$$ New conditions are more convenient and can be rewritten to: $$b^2<c<\frac{2b}{1+b}.$$ Let's devide the condition by $b$ and the hypothesis by $b^2$, the problem is equivalen to: $$b<\frac{c}{b}<\frac{2}{1+b}\Rightarrow1+\Big(\frac{c}{b}\Big)^2\le\frac{c}{b}+\frac{1}{b^2}.$$ Let's introduce $r:=\frac{c}{b}$. The problem is equivalent to: $$b<r<\frac{2}{1+b}\Rightarrow1-r+r^2\le\frac{1}{b^2}.$$ The condition implies: $$b<\frac{2}{1+b}\Rightarrow b+b^2<2\Rightarrow\Big(b+\frac{1}{2}\Big)^2<\frac{9}{4}\Rightarrow b+\frac{1}{2}<\frac{3}{2}\Rightarrow b<1.$$ Returning the limitation $b>0$ together with the $\frac{2}{1+b}$ being decreasing with the supremum at $b=0$ we get: $$0<b<1\land0<r<2.$$ The case $r\le1$ is easy: $$1-r+r^2=\frac{3}{4}+\Big(\frac{1}{2}-r\Big)^2\le\frac{3}{4}+\max_{0<r\le1}\Big|\frac{1}{2}-r\Big|^2=\frac{3}{4}+\Big(\frac{1}{2}\Big)^2=1=\frac{1}{1}<\frac{1}{b^2}.$$ For $r>1$ we have $$\frac{d}{dr}(1-r+r^2)=2r-1>1>0,$$ so the expression $1-r+r^2$ increases with $r$ and we have $$1-r+r^2<1-\frac{2}{1+b}+\Big(\frac{2}{1+b}\Big)^2=\frac{3+b^2}{(1+b)^2}=\frac{1}{b^2}\frac{(3+b^2)b^2}{(1+b)^2}=$$ $$\frac{1}{b^2}\frac{b^2+2b^2+b^4}{(1+b)^2}<\frac{1}{b^2}\frac{1+2b+b^2}{(1+b)^2}=\frac{1}{b^2}.$$

Answer 2

Another solution.

First, we find $c<\frac{2ab}{a+b}\leq \frac{2\left(\frac{a+b}{2}\right)^2}{a+b}=\frac{a+b}{2}$ by applying AM-GM to $ab$. We also find $b<\sqrt{ac}\leq \frac{a+c}{2}$.

Now if $(b+c)\leq a$, this implies $(b+c)^2\leq a^2 \Rightarrow b^2+c^2-a^2\leq -2bc<bc$, and we are done.

If not, then $b+c>a$ we'll have a triangle $\triangle ABC$. By the law of cosine,

$$a^2 = b^2+c^2-2bc\cos{A} \Rightarrow b^2+c^2-a^2=2bc\cos{A}$$ we only need to consider cases where $\angle A<60^{\circ}$ which would make our statement false.

Now $\angle A<60^{\circ}$ implies $a$ can't be the longest side, and the condition $b<\sqrt{ac}$ implies $b$ can't be the longest side. Therefore $c$ must be the longest side. However $c<\frac{a+b}{2}$, so that contradicts the fact we have a triangle.

Answer 3

Using $ac - b^2 > 0$ and $2ab - (a + b)c > 0$, we have \begin{align*} &bc - (b^2 + c^2 - a^2)\\ >\,&bc - (b^2 + c^2 - a^2) - (ac - b^2) - [2ab - (a + b)c]\\ =\,& a^2 - 2ab + 2bc - c^2\\ =\,&(a - c)(a + c - 2b)\\ >\,& 0 \end{align*} where we have used $a > c$ and $a + c > 2b$.
(Explanations: By AM-GM, we have $2b < 2\sqrt{ac} \le a+c$. Also, by AM-GM, we have $$c < \frac{2ab}{a + b} \le \frac{2ab}{2\sqrt{ab}} = \sqrt{ab} < \sqrt{a \sqrt{ac}}$$ which results in $c < a$.)

We are done.

Answer 4

Let $a<\sqrt{b^2-bc+c^2}.$

Thus, $$b^2<ac<c\sqrt{b^2-bc+c^2},$$ which gives $$(b-c)(b^2(b+c)+c^3)<0$$ or $$b<c.$$ Also, since $$a(2b-c)>bc,$$ we obtain: $$2b-c>0$$ and $$\sqrt{b^2-bc+c^2}>a>\frac{bc}{2b-c},$$ which gives $$(b-c)(4b^3-4b^2c+4bc^2-c^3)>0$$ or $$4b^3-4b^2c+4bc^2-c^3<0.$$ But $$4b^3-4b^2c+4bc^2-c^3=b(2b-c)^2+c^2(2b-c)+c^2b>0,$$ which is a contradiction.