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I have two similar looking questions.

(1) Prove that in triangle $ABC$, $$\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}.$$

(2) If $\Delta ABC$ is an acute angled, then prove that
$$\cos^2A+\cos^2B+\cos^2C<\frac{3}{2}$$

If I apply Jensen's inequality, then $\cos^2x$ is a concave function, because its second derivative is $-2\cos 2x$ and with it being concave function $$\cos^2A+\cos^2B+\cos^2C\leq\frac{3}{4}$$ which is not there in the question.How will we prove both of these questions. I have some intuition that in the second question, as $ABC$ is an acute angled triangle,this has something to do.

Please guide me in the right direction. Thank you.

Community
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user1442
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    $-2\cos(2x)$ is not always non-negative, even for angles of a triangle. – Quang Hoang Sep 03 '15 at 12:44
  • It isn't true that $cos^2(A)+cos^2(B)+cos^2(C)≤\frac 34$. For the right isosceles triangle you get $1$, which I believe is the upper bound in the acute case. – lulu Sep 03 '15 at 12:53

5 Answers5

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$$y=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$

Now $\cos^2A-\sin^2B=\cos(A+B)\cos(A-B)=-\cos C\cos(A-B)$

$$\implies\cos^2C-\cos C\cos(A-B)+1-y=0$$ which is a Quadratic Equation in $\cos C$

$$\implies\cos^2(A-B)-4(1-y)\ge0\iff4y\ge4-\cos^2(A-B)=3+\sin^2(A-B)$$

$$\implies4y\ge3$$

The equality occurs if $\sin(A-B)=0\implies A=B\ \ \ \ (1)$ and $\cos C=\dfrac12\implies C=\dfrac\pi3\ \ \ \ (2)$

$(1),(2)\implies A=B=C$

$$\cos^2A+\cos^2B+\cos^2C=1+\cos^2C-\cos C\cos(A-B)$$

$$=1+\cos C[\cos C-\cos(A-B)] $$

$$=1-\cos C[\cos(A+B)+\cos(A-B)] $$

$$=1-2\cos A\cos B\cos C<1$$ if $0<A,B,C<\dfrac\pi2$

lab bhattacharjee
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Multiplying by $4R^2$ and exploiting the sine theorem we get:

$$ 4R^2\sum_{cyc}\cos^2 A = 12R^2-(a^2+b^2+c^2)$$ hence the first inequality is equivalent to the trivial $OH^2\geq 0$, where $O$ is the circumcenter and $H$ is the orthocenter. On the other hand, if $ABC$ is an acute-angled triangle we have that $H$ lies inside $ABC$, hence $OH^2< R^2$ and the second inequality follows.

Jack D'Aurizio
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Eliminate $C$ from $ A+B+C= \pi$

Next, recognize the bounds of $ \cos A, \cos B.$

Narasimham
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Try the first one. Transform the triangle using linear transformations(thereby keeping angles invariant) to A(0,a),B(-b,0),C(c,0). What we are after is bb/(aa+bb) +cc/(aa+cc)+(aa/(sqrt(aa+cc)sqrt(aa+bb))-cb/(sqrt(aa+cc).sqrt(aa+bb)))^2>=3/4. Simplifying then we are after 4(3bbcc+aaaa+bbaa+aacc)>=3(aaaa+aabb+aacc+bbcc). This is clear.

Adelafif
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Another way.

(1) We need to prove that $$1+\cos2\alpha+1+\cos2\beta+2\cos^2\gamma\geq\frac{3}{2}$$ or $$4\cos(\alpha+\beta)\cos(\alpha-\beta)+4\cos^2\gamma+1\geq0$$ or $$\left(2\cos\gamma-\cos(\alpha-\beta)\right)^2+\sin^2(\alpha-\beta)\geq0.$$ (2) Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$\sum_{cyc}\frac{x^2}{2\cdot\frac{x+y}{2}\cdot\frac{x+z}{2}}<\frac{3}{2}$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+2xyz)>0,$$ which is true because by the Schur's inequality even $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0.$$

Michael Rozenberg
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