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Tag Archives: Inequality

Trigonometric inequality on a triangle

Let ABC be a triangle. Prove that

    \[\frac{1}{\sin^2 A} + \frac{1}{\sin^2 B} + \frac{1}{\sin^2 C} \geq 4\]

Solution

The function \displaystyle f(x) = \frac{1}{\sin^2 x } is convex, thus:

    \begin{align*} \frac{1}{3} \sum f(A) &= \frac{1}{3} \sum \frac{1}{\sin^2 A}\\ &\geq f\left ( \frac{1}{3} \sum A \right ) \\ &= \frac{1}{\sin^2 \left ( \frac{1}{3} \sum A \right )} \\ &= \frac{1}{\sin^2 \frac{\pi}{3}} \\ &= \frac{4}{3} \end{align*}

The result follows. In fact, something a bit stronger holds. Let r denote the inradius , R the circumradius and s the semiperimeter. Then,

    \[\frac{1} {\sin ^2 A} + \frac{1} {\sin^2 B} + \frac{1} {\sin^2 C} \geq \frac{2R}{r}\]

Indeed,

    \begin{align*} \sum \frac{1}{\sin^2 A} &= 4R^2\sum \frac{1}{a^2} \\ &\geq 4R^2 \sum \frac{1}{ab}\\ &=\frac{4R^2 \left ( a+b+c \right )}{abc}\\ &= \frac{8R^2s}{abc} \\ &= 8R^2 \cdot \frac{{\rm E}}{r} \cdot \frac{1}{4R{\rm E}} \\ &= \frac{2R}{r} \end{align*}

in view of the known identities

(1)   \begin{equation*} E = rs \end{equation*}

(2)   \begin{equation*} 4R\mathrm{E} = abc \end{equation*}

(3)   \begin{equation*} R \geq 2 r \end{equation*}

 

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Inequality involving area of a triangle

Show that in any triangle ABC with area \mathcal{A} the following holds:

    \[\frac{1}{2} \left ( \frac{\sin A + \sin B + \sin C}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right ) \leq \mathcal{A}\]

Solution

Let s is the semiperimeter , R the circumradius and r the inradius. From the law of sines we find

(1)   \begin{equation*}  \sin A + \sin B + \sin C = \frac{s}{R} \end{equation*}

as well as

(2)   \begin{equation*} \mathcal{A} = rs \end{equation*}

Now,

    \[\frac{1}{2\mathcal{A}} \sum \sin A \leq \sum \frac{1}{a^2}\]

Substitute the preceding equalities into the last inequality  and simplifying we obtain

    \[\frac{1}{2rR} \leq \sum \frac{1}{a^2}\]

From

    \[\mathcal{A} = rs = \frac{abc}{4R}\]

the last inequality becomes

    \[\frac{a+b+c}{abc} = \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab} \leq \frac{1}{a^2}+\frac{1}{b^2} + \frac{1}{c^2}\]

which is true by the rearrangement inequality.

 

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Trigonometric inequality on an acute triangle

Prove that in any acute triangle ABC the following inequality holds:

    \[\frac{\sin A \sin B}{\cos C} + \frac{\sin B \sin C}{\cos A} + \frac{\sin C \sin A}{\cos B} \geq \frac{9}{2}\]

Solution

Since A+B+C = \pi it holds that

(1)   \begin{equation*} \cos C = -\cos (A +B ) = -\cos A \cos B + \sin A \sin B \end{equation*}

and thus

(2)   \begin{equation*} \frac{\sin A \sin B}{\cos C} = 1 + \frac{\cos A \cos B}{\cos C} = 1 + \frac{\tan C}{\tan A + \tan B} \end{equation*}

Using Nesbitt’s inequality we see that

    \begin{align*} \sum \frac{\sin A \sin B}{\cos C} &= \sum \left ( 1 + \frac{\tan C}{\tan A + \tan B} \right ) \\ &=3 + \sum \frac{\tan C}{\tan A + \tan B} \\ &\geq 3 + \frac{3}{2} \\ &= \frac{9}{2} \end{align*}

Equality holds if-f A=B=C=\frac{\pi}{3}.

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Trigonometric inequality

Prove that the following inequality holds in any triangle:

    \[\left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 + \left ( \sin \frac{B}{2} + \sin \frac{C}{2} \right )^2 + \left ( \sin \frac{C}{2} + \sin \frac{A}{2} \right )^2 \leq 3\]

Solution

Let s denote the semiperimeter of the triangle. Using the cosine theorem we have that

    \[c^2=a^2+b^2-2ab \cos C = \left ( a-b \right )^2 + 4ab \sin^2 \frac{C}{2}\]

from which it follows that

(1)   \begin{equation*} \sin^2 \frac{C}{2} = \frac{\left ( s-a \right )\left ( s-b \right )}{ab}\end{equation*}

(2)   \begin{equation*} \sin^2 \frac{A}{2} = \frac{\left ( s-b \right )\left ( s-c \right )}{bc} \end{equation*}

(3)   \begin{equation*} \sin^2 \frac{B}{2} = \frac{\left ( s-a \right )\left ( s-c \right )}{ac} \end{equation*}

Thus, by Cauchy’s inequality we have:

    \begin{align*} \left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 &= \frac{s-c}{c} \left ( \sqrt{s-b} \cdot \frac{1}{\sqrt{b}} +\sqrt{s-a} \cdot \frac{1}{\sqrt{a}} \right )^2 \\ &\leq \frac{s-c}{c} \cdot \left ( \frac{1}{a} + \frac{1}{b} \right )\cdot \left ( s-a+s-b \right ) \\ &= \frac{\left ( a+b-c \right )\left ( a+b \right )c}{2abc}\\ &= \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right ) +2abc}{2abc} \\ &= 1 + \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right )}{2abc} \end{align*}

Hence,

    \begin{align*} \sum \left ( \sin \frac{A}{2} + \sin \frac{B}{2} \right )^2 &\leq 3 + \sum \frac{\left ( a^2+b^2 \right )c -c^2\left ( a+b \right )}{2abc} \\ &=3 + \frac{1}{2abc} \left ( \sum \left ( a^2+b^2 \right )c - \sum c^2 \left ( a+b \right ) \right ) \\ &= 3 \end{align*}

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Inequality of a function

Let f:[0, 1] \rightarrow \mathbb{R} be a differentiable function with continuous derivative. Prove that:

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1} \left | f(x) \right | \, \mathrm{d}x + \frac{1}{2} \int_{0}^{1} \left | f'(x) \right | \, \mathrm{d}x\]

Solution

For 0 \leq x \leq \frac{1}{2} it holds that

    \[f\left ( \frac{1}{2} \right ) -f(x) = \int_{x}^{1/2} f'(t)\, \mathrm{d}t\]

Taking absolute values and using basic properties of the integral we get

    \[\left | f\left ( \frac{1}{2} \right ) \right | \leq \left | f(x) \right | + \int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t\]

Integrating we have:

(1)   \begin{equation*}   \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{0}^{1/2}\left | f(x) \right | + \frac{1}{2}\int_{0}^{1/2} \left | f'(t) \right | \, \mathrm{d}t  \end{equation*}

Working similarly on \left[\frac{1}{2} , 1 \right] we get

(2)   \begin{equation*} \frac{1}{2}\left | f\left ( \frac{1}{2} \right ) \right | \leq \int_{1/2}^{1}\left | f(x) \right | + \frac{1}{2}\int_{1/2}^{1} \left | f'(t) \right | \, \mathrm{d}t    \end{equation*}

Adding equations (1), (2) we get the result.

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