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Tag Archives: Inequality

Root inequality

Let x, y, z>0 such that \sqrt{xy}+\sqrt{yz}+\sqrt{yz}=1. Prove that

    \[\frac{x^{2}}{x+y}+\frac{y^{2}}{y+z}+\frac{z^{2}}{z+x}\geq \frac{1}{2}\]

Solution

It follows from Cauchy – Schwartz that

    \begin{align*} \frac{x^{2}}{x+y}+\frac{y^{2}}{y+z}+\frac{z^{2}}{z+x} &\geq \frac{(x+y+z)^2}{2(x+y+z)}\\ &=\frac{x+y+z}{2}\\ &\geq \frac{\sqrt{xy}+\sqrt{yz}+\sqrt{yz}}{2}\\ &= \frac{1}{2} \end{align*}

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Double inequality involving matrix

Prove that

    \[\max_{1\leq j\leq p}\sqrt{\sum_{i=1}^{q}a^2_{ij}}\leq \left \| \begin{pmatrix} a_{11} &\cdots &a_{1p} \\ \vdots& \ddots &\vdots \\ a_{q1}&\cdots &a_{qp} \end{pmatrix} \right \|_2 \leq \sqrt{\sum_{i=1}^{q}\sum_{j=1}^{p}a^2_{ij}}\]

Solution

Fix j. Apply the matrix A on e_j thus:

    \[\Vert Ae_j\Vert _2 \le \Vert A\Vert _2 \Vert e_j\Vert _2= \Vert A\Vert _2\]

Since Ae_j is exactly the j – th column of A the previous equality can be rerwritten as

    \[\sqrt{\sum_{i=1}^{q}a^2_{ij}}\leq \Vert A \Vert _2\]

Since this holds for all j=1, 2, \dots, n we get \max and the left inequality follows.

For a random unit vector x=(x_1, x_2, \dots, x_n) the i coordinate of the vector Ax is \sum \limits_j a_{ij}x_j. It follows from Cauchy – Schwartz that

    \[\left |\sum _ja_{ij}x_j\right |^2 \le \sum _j|a_{ij}|^2\sum _j|x_j|^2 = \sum _j|a_{ij}|^2\]

Summing over all i‘s till we find \Vert Ax \Vert _2 we conclude that, for every unit vector x, it holds that \Vert Ax \Vert _2 is less than the right hand side. Taking supremum with respect to all x the right hand side inequality follows.

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Homogeneity of inequality

Let x, y, z>0. Prove that:

    \[\frac{y^3z}{x^2(xy+z^2)} +\frac{z^3x}{y^2(zy+x^2)} +\frac{x^3y}{z^2(xz+y^2)} \geq \frac{3}{2}\]

Solution

Due to homogeneity we may assume xyz=1. Thus there exist positive a, b, c such that

    \[x=\frac{a}{b}\quad , \quad y=\frac{b}{c}\quad ,\quad z=\frac{c}{a}\]

Hence,

    \begin{align*} \sum \frac{y^3z}{x^2(xy+z^2)}  &= \frac{a^5}{bc(b^3+c^3)}+\frac{b^5}{ca(c^3+a^3)}+\\ & \quad \quad \quad +\frac{c^5}{ab(a^3+b^3)}\\ &=\frac{a^6}{abc(b^3+c^3)}+\frac{b^6}{abc(c^3+a^3)}+\\ &\quad \quad \quad + \frac{c^6}{abc(a^3+b^3)} \\ &\geq \frac{(a^3+b^3+c^3)^2}{2abc(a^3+b^3+c^3)}\\ &=\frac{1}{2}\frac{a^3+b^3+c^3}{abc}\\ &\geq \frac{3}{2} \end{align*}

 

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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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