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Tag Archives: Inequality

Root inequality

Let a, b, c be positive real numbers such that a+b+c=3. Prove that

    \[\sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} \leq \frac{3}{2} \sqrt[4]{\frac{1}{abc}}\]


Due to the AM – GM we have that

(1)   \begin{equation*} a^2+3 \geq 4\sqrt{a} \end{equation*}


(2)   \begin{equation*} 3b + c \geq 4 \sqrt[4]{b^3c} \end{equation*}


    \begin{align*} \sum \sqrt{\frac{b}{a^2+3}}  &\leq \sqrt{\frac{b}{4\sqrt{a}}} + \sqrt{\frac{c}{4\sqrt{b}}} + \sqrt{\frac{a}{4\sqrt{c}}} \\ &=\frac{1}{2\sqrt[4]{abc}} \left ( \sqrt[4]{b^3c} + \sqrt[4]{c^3a} + \sqrt[4]{a^3b} \right ) \\ &\leq \frac{1}{2\sqrt[4]{abc}} \left ( \frac{3b+c}{4} + \frac{3c+a}{4} + \frac{3a+b}{4} \right ) \\ &=\frac{3}{2} \sqrt[4]{\frac{1}{abc}} \end{align*}

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Inequality of a triangle

Let ABC be a triangle and denote a, b, c the lengths of the sides BC , CA and AB respectively. If abc \geq 1 then prove that

    \[\sqrt{\frac{\sin A}{a^3+b^6+c^6}} + \sqrt{\frac{\sin B}{b^3+c^6+a^6}} + \sqrt{\frac{\sin C}{c^3 + a^6+b^6}} \leq \sqrt[4]{\frac{27}{4}}\]


Applying Cauchy’s inequality to the vectors

    \[\mathbf{u} = \left ( \sqrt{\sin A} , \sqrt{\sin B} , \sqrt{\sin C} \right )\]


    \[\mathbf{v} = \left ( \sqrt{\frac{1}{a^3+b^6+c^6}} , \sqrt{\frac{1}{b^3+c^6+a^6}} , \sqrt{\frac{1}{c^3+a^6+b^6}} \right )\]

we get that

    \begin{align*} \left ( \sum \sqrt{\frac{\sin A}{a^3+b^6+c^6}} \right )^2 &\leq \left ( \sum \sin A \right ) \left ( \sum \frac{1}{a^3+b^6+c^6} \right ) \\ &\leq \frac{3\sqrt{3}}{2} \sum \frac{1}{a^3+b^6+c^6} \end{align*}

due to the well known fact

(1)   \begin{equation*} \sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2} \end{equation*}

It suffices to prove that \displaystyle \sum \frac{1}{a^3+b^6+c^6} \leq 1. Let x, y be two positive real numbers. Then,

(2)   \begin{equation*} \left ( x-y \right )^2\left ( x^4+x^3y + x^2y^2 + xy^3+ y^4 \right )\geq 0\Leftrightarrow x^6+y^6 \geq xy \left ( x^4+y^4 \right ) \end{equation*}

On the other hand if xyz \geq 1 then x + y + z \geq 3\sqrt[3]{xyz} =3 and

(3)   \begin{equation*} x^4+y^4+z^4 \geq \frac{\left ( x+y+z \right )^4}{27}  \end{equation*}

Then it follows that

    \begin{align*} \sum \frac{1}{a^3+b^6+c^6} &\leq \sum \frac{1}{ab\left ( a^4+b^4 \right )+c^3} \\ &\leq \sum \frac{1}{\frac{a^4+b^4}{c}+ c^3}\\ &= \sum \frac{c}{a^4+b^4+c^4} \\ &=\frac{a+b+c}{a^4+b^4+c^4}\\ &\leq \frac{27}{\left ( a+b+c \right )^3}\\ &\leq 1 \end{align*}

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Root inequality

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

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


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

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


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:

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


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

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


    \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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