Introduction to Inequalities

After subtracting \(2ab\) from both sides the inequality becomes equivalent to \((a-b)^2\geq 0\), which is true according to Theorem 1.

If we add the inequalities \(a^2+b^2\geq 2ab\), \(b^2+c^2\geq 2bc\), and \(c^2+a^2\geq 2ca\) we get \(2a^2+2b^2+2c^2\geq 2ab+2bc+2ca\), which is equivalent to what we are asked to prove.

Recall that \(x^2+y^2\geq 2xy\), where the equality holds if and only if \(x=y\). Applying this inequality to the pairs of numbers \(\left(a/2,b\right)\), \(\left(a/2,c\right)\), and \(\left(a/2,d\right)\) yields: \begin{eqnarray*} \frac{a^2}4+b^2\geq ab,\;\; \frac{a^2}4+c^2\geq ac,\;\; \frac{a^2}4+d^2\geq ad. \end{eqnarray*} Note also that \(a^2/4 > 0\). Adding these four inequalities gives us \(a^2+b^2+c^2+d^2\geq a(b+c+d)\). Equality can hold only if all the inequalities were equalities, i.e. \(a^2=0\), \(a/2=b\), \(a/2=c\), \(a/2=d\). Hence \(a=b=c=d=0\) is the only solution of the given equation.

If we apply the inequality \(x^2+y^2\geq 2xy\) to the numbers \(\displaystyle x=\frac{ab}c\) and \(\displaystyle y=\frac{bc}a\) we get \begin{eqnarray*} \frac{a^2b^2}{c^2}+ \frac{b^2c^2}{a^2} \geq 2b^2.\quad\quad\quad\quad\quad(1) \end{eqnarray*} Similarly we get \begin{eqnarray*} \frac{b^2c^2}{a^2}+ \frac{c^2a^2}{b^2} \geq 2c^2,\; \mbox{and} \quad\quad\quad\quad\quad(2) \end{eqnarray*} \begin{eqnarray*} \frac{c^2a^2}{b^2}+ \frac{a^2b^2}{c^2} \geq 2a^2.\quad\quad\quad\quad\quad(3) \end{eqnarray*} Summing up (1), (2), and (3) gives \(2\left(\frac{a^2b^2}{c^2}+ \frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\right) \geq 2(a^2+b^2+c^2)=2\), hence \(S\geq 1\). The equality holds if and only if \(\displaystyle\frac{ab}c=\frac{bc}a=\frac{ca}b\), i.e. \(a=b=c = \displaystyle\frac1{\sqrt 3}\).

Using the inequality \(a+b\geq2\sqrt{ab}\) we get \(\frac1{1-x^2}+\frac1{1-y^2}\geq \frac2{\sqrt{(1-x^2)(1-y^2)}}\). Now we notice that \((1-x^2)(1-y^2)=1+x^2y^2-x^2-y^2\leq 1+x^2y^2-2xy=(1-xy)^2\) which implies \( \frac2{\sqrt{(1-x^2)(1-y^2)}}\geq \frac2{1-xy}\) and this completes the proof.

The given inequality is equivalent to \(a^3+b^3-a^2b-ab^2\geq 0\). The left-hand side can be transformed in the following way: \(a^3+b^3-a^2b-ab^2=a^2(a-b)+b^2(b-a)=(a-b)(a^2-b^2)=(a-b)^2(a+b)\geq 0\).

Let us introduce the substitution \(x=\frac{a^2}{1+a^2}\), \(y=\frac{b^2}{1+b^2}\), \(z=\frac{c^2}{1+c^2}\). The numbers \(x\), \(y\), and \(z\) are positive real numbers smaller than \(1\) that satisfy \(x+y+z=1\). From the equation \(x=\frac{a^2}{1+a^2}\) we immediately obtain \(a^2=\frac{x}{1-x}=\frac{x}{y+z}\). In a similar way we derive \(b^2=\frac{y}{z+x}\) and \(c^2=\frac{z}{x+y}\). The required inequality is equivalent to \[xyz\leq \frac18(y+z)(z+x)(x+y).\] The last equality follows directly from \(x+y\geq 2\sqrt{xy}\), \(y+z\geq 2\sqrt{yz}\), and \(z+x\geq 2\sqrt {zx}\) The equality holds if and only if \(x=y=z\), i.e. \(a^2=b^2=c^2=\frac12\).

The trick we use here is the substitution \(x=b+c\), \(y=c+a\), \(z=a+b\). This will allow us to transform the inequality into another one that doesn’t have ugly denominators. We first recover our variables \(a\), \(b\), \(c\) as: \(a=\frac{y+z-x}2\), \(b=\frac{z+x-y}2\), \(c=\frac{x+y-z}2\), and then we see that the required inequality is equivalent to \begin{eqnarray*}\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}&\geq& \frac32\\ \Leftrightarrow \;\; \left(\frac xy+\frac yx\right)+\left(\frac yz+\frac zy\right)+\left(\frac zx+\frac xz\right)\geq 6.\end{eqnarray*} The last inequality is true because \(\frac xy+\frac yx\geq 2\sqrt{\frac xy\cdot \frac yx}=2\) and similar inequalities hold for the other two terms. The equality holds if and only if \(x=y=z\), i.e. if and only if \(a=b=c\).

Let us multiply both sides of \(a^2+b^2+c^2-ab-bc-ca\geq 0\) by \((a+b+c)\). After some calculation we see that the left side transforms to \(a^3+b^3+c^3-3abc\) and the inequality becomes \(a^3+b^3+c^3-3abc\geq 0\) which is equivalent to the required inequality.

Applying the inequality from theorem 3 impies that \[\frac13(2a^3+b^3)=\frac13(a^3+a^3+b^3)\geq a \cdot a \cdot b =a^2b.\] Multiplying both sides by \(3\) gives the desired result.

Using the result of Problem 8 we get \(2a^3+b^3\geq 3a^2b\). Adding this with analogous inequalities \(2b^3+c^3\geq 3b^2c\) and \(2c^3+a^3\geq 3c^2a\) gives us \(3(a^3+b^3+c^3)\geq 3(a^2b+b^2c+c^2a)\). After dividing by \(3\) we get the desired inequality.

Denote by \(L\) the left-hand side of the required inequality. If we add the first and the third summand of \(L\) we get \[\frac{a}{b+c}+\frac{c}{d+a} =\frac{a^2+c^2+ad+bc}{(b+c)(a+d)}. \] We will bound the denominator of the last fraction using the inequality \(xy\leq (x+y)^2/4\) for appropriate \(x\) and \(y\). For \(x=b+c\) and \(y=a+d\) we get \((b+c)(a+d)\leq(a+b+c+d)^2/4\). The equality holds if and only if \(a+d=b+c\). Therefore \[\frac{a}{b+c}+\frac{c}{d+a}\geq 4\frac{a^2+c^2+ad+bc}{(a+b+c+d)^2}.\] Similarly \(\frac{b}{c+d}+\frac{d}{a+b}\geq 4\frac{b^2+d^2+ab+cd} {(a+b+c+d)^2}\) (with the equality if and only if \(a+b=c+d\)) implying \begin{eqnarray*} && \frac a{b+c}+\frac b{c+d}+ \frac c{d+a}+ \frac d{a+b} \\ &\geq &4\frac{a^2+b^2+c^2+d^2+ad+bc+ab+cd}{(a+b+c+d)^2} \\ \nonumber &= & 4\frac{a^2+b^2+c^2+d^2+(a+c)(b+d)}{[(a+c)+(b+d)]^2}. \end{eqnarray*} In order to solve the problem it is now enough to prove that \begin{eqnarray*} 2\frac{a^2+b^2+c^2+d^2+(a+c)(b+d)}{[(a+c)+(b+d)]^2}\geq 1. \end{eqnarray*} After multiplying both sides of the last inequality by \([(a+c)+(b+d)]^2=(a+c)^2+(b+d)^2\) it becomes equivalent to \(2(a^2+b^2+c^2+d^2)\geq (a+c)^2+(b+d)^2=a^2+b^2+c^2+d^2+2ac+2bd.\) It is easy to see that the last inequality holds because many terms will cancel and the remaining inequality is the consequence of \(a^2+c^2\geq 2ac\) and \(b^2+d^2\geq 2bc\). The equality holds if and only if \(a=c\) and \(b=d\).

We first notice that \[\frac{a^3-b^3}{a^2+ab+b^2}+ \frac{b^3-c^3}{b^2+bc+c^2}+ \frac{c^3-a^3}{c^2+ca+a^2} =0.\] Hence it is enough to prove that \[\frac{a^3+b^3}{a^2+ab+b^2}+ \frac{b^3+c^3}{b^2+bc+c^2}+ \frac{c^3+a^3}{c^2+ca+a^2} \geq \frac{2(a+b+c)}3.\] However since \(3(a^2-ab+b^2)\geq a^2+ab+b^2\), \[\frac{a^3+b^3}{a^2+ab+b^2}= (a+b)\frac{a^2-ab+b^2}{a^2+ab+b^2}\geq \frac{a+b}3.\] The equality holds if and only if \(a=b=c\).

Second solution. First we prove that \begin{eqnarray*} \frac{a^3}{a^2+ab+b^2}\geq \frac{2a-b}3.\quad\quad\quad\quad\quad(1)\end{eqnarray*} Indeed after multiplying we get that the inequality is equivalent to \(a^3+b^3\geq ab(a+b)\), or \((a+b)(a-b)^2\geq 0\) which is true. After adding (1) with two similar inequalities we get the result.

We will prove the statement using the induction. Let us first prove the case \(n=2\). We need to prove that \begin{eqnarray*} \frac{ab}{a+b}+ \frac{cd}{c+d} \leq \frac{(a+c)(b+d)}{a+b+c+d}. \end{eqnarray*} Multiply both sides of the previous inequality by \((a+b)(c+d)(a+b+c+d)\). Many things cancel out and what remains is to verify the inequality \(4abcd\leq a^2d^2+b^2c^2\) which is true because it is equivalent to \(0\leq (ad-bc)^2\). The equality holds if and only if \(ad=bc\), or \(\frac ab=\frac cd\).

If we now assume that the inequality holds for some \(n\) then, \begin{eqnarray*} \frac{a_1b_1}{a_1+b_1}+\cdots+\frac{a_nb_n}{a_n+b_n}+\frac{a_{n+1}b_{n+1}}{a_{n+1}+b_{n+1}}&\leq& \frac{(a_1+\cdots+a_n)(b_1+\cdots+b_n)}{a_1+\cdots+a_n+b_1+\cdots+b_n}+\frac{a_{n+1}b_{n+1}}{a_{n+1}+b_{n+1}}\\&\leq & \frac{(a_1+\cdots+a_{n+1})(b_1+\cdots+b_{n+1})}{a_1+\cdots+a_{n+1}+b_1+\cdots+b_{n+1}}. \end{eqnarray*}

By induction it is easy to prove more general statement: \[\sqrt{a_1b_1}+\sqrt{a_2b_2}+\cdots+\sqrt{a_nb_n}\leq\sqrt{(a_1+\cdots+a_n)(b_1+\cdots+b_n)}.\] The case \(n=2\) is easy to verify. Indeed \(\sqrt{ab}+\sqrt{cd}\leq\sqrt{(a+c)(b+d)}\) is equivalent to \(ab+cd+2\sqrt{abcd}\leq ab+ad+bc+cd\), i.e. \(2\sqrt{abcd}\leq ad+bc\) which follows directly from the AMGM inequality. If the statement is true for \(n\) then \begin{eqnarray*} \sqrt{a_1b_1}+\cdots+\sqrt{a_nb_n}+\sqrt{a_{n+1}b_{n+1}}&\leq &\sqrt{(a_1+\cdots+a_n)(b_1+\cdots+b_n)}+\sqrt{a_{n+1}b_{n+1}}\\ &\leq &\sqrt{(a_1+\cdots+a_{n+1})(b_1+\cdots+b_{n+1})}. \end{eqnarray*}

The given inequality follows from \(\frac{5a^3-ab^2}{a+b}\geq 3a^2-b^2\). In order to prove the last relation we notice that it is equivalent to \(2a^3+b^3\geq 3a^2b\). This inequality follows from Theorem 3.

Define \(x_0=0\), \(x_{n+1}=1\), \(y_k=x_{k+1}-x_k\) for \(k=0,1,2,\dots, n\). Then \(y_0=y_n=0\). Assume the contrary, i.e. that \(|x_{j+1}+x_{j-1}-2x_j|=|y_{j}-y_{j-1}| < \frac4{n^2}\) for each \(j\in\{1,2,\dots, n\}\). Adding the inequalities \(y_j-y_{j-1} < \frac4{n^2}\) for \(j=1,2,\dots, k\) we get \(y_k < \frac{4k}{n^2}\). Similarly, by adding \(y_j-y_{j+1} < \frac{4}{n^2}\) for \(j=k,k+1,\dots, n-1\) we get \(y_k < \frac{4(n-k)}{n^2}\). Consider the following two cases:

• Case 1. If \(n\) odd then \begin{eqnarray*}1&=& y_0+y_1+\cdots+ y_n=\left(y_1+\cdots + y_{\frac{n-1}2}\right)+\left(y_{\frac{n+1}2}+\cdots + y_{n}\right)\\&< & \frac{4}{n^2}\cdot \frac12\cdot\left(\frac{n-1}2\cdot\frac{n+1}2\cdot 2\right) =\frac{n^2-1}{n^2},\end{eqnarray*} a contradiction.

• Case 2. If \(n\) is even, we get a contradiction as well since \begin{eqnarray*}1&=& y_0+y_1+\cdots+ y_n=\left(y_1+\cdots + y_{\frac{n}2-1}\right)+\left(y_{\frac{n}2}+\cdots + y_{n-1}\right)\\&< & \frac{4}{n^2}\cdot \frac12\cdot\left(\left(\frac{n}2-1\right)\cdot \frac{n}2+\frac n2\cdot \left(\frac n2+1 \right) \right)=1.\end{eqnarray*}