Problem 1. For all positive real numbers \( a, b, c \), prove that

 

\[
\frac{(a^2 + bc)^2}{b + c} + \frac{(b^2 + ca)^2}{c + a} + \frac{(c^2 + ab)^2}{a + b} \geq \frac{2abc(a + b + c)^2}{ab + bc + ca}.
\]

Solution 1
Step 1: Homogeneity and Assumption
First, we notice that the inequality is homogeneous. This means that all the terms are of the same degree. Therefore, we can assume \( abc = 1 \). This simplifies the problem because in homogeneous inequalities, such assumptions are often helpful.

Step 2: Using the AM-GM Inequality

Next, we apply the \textbf{AM-GM inequality} (Arithmetic-Geometric Mean inequality). This inequality is always expressed as:

\[
\frac{a + b}{2} \geq \sqrt{ab}
\]
In this case, we apply the AM-GM inequality as follows:

\[
\sum_{\text{cyc}} \frac{(a^2 + bc)^2}{b+c} \geq \sum_{\text{cyc}} \frac{4a}{b+c}
\]
So, by applying AM-GM, we get the first inequality:

\[
\sum_{\text{cyc}} \frac{(a^2 + bc)^2}{b+c} \geq \sum_{\text{cyc}} \frac{4a}{b+c}
\]

Step 3: Second Inequality

In this step, we simplify the expression on the right-hand side further. The previously obtained sum:

\[
\sum_{\text{cyc}} \frac{4a}{b+c}
\]
is rewritten as:

\[
\sum_{\text{cyc}} \frac{4a^2}{ab+ac}
\]
Here, the terms are reorganized and simplified.



Step 4: Conclusion

 

Finally, we arrive at the inequality:

\[
\sum_{\text{cyc}} \frac{4a^2}{ab+ac} \geq \frac{2(a + b + c)^2}{ab + bc + ca}
\]
This inequality can then be written as:

\[
\frac{2(a + b + c)^2}{ab + bc + ca} = \frac{2abc(a + b + c)^2}{ab + bc + ca}
\]
Thus, the two sides are equal, and the desired result is achieved.

Solution 2

Apply Cauchy-Schwarz inequality.
$$((b+c)+(c+a)+(a+b))\cdot \left(\dfrac{(a^2+bc)^2}{b+c}+\dfrac{(b^2+ca)^2}{c+a}+\dfrac{(c^2+ab)^2}{a+b}\right)\geq (a^2+b^2+c^2+ab+bc+ca)^2$$
Lemma. $3(a^2+b^2+c^2+ab+bc+ca)\geq 2(a+b+c)^2$

Proof. This is equivalent to well known $a^2+b^2+c^2\geq ab+bc+ca$ when expanded out.

Then we have
$$\dfrac{(a^2+bc)^2}{b+c}+\dfrac{(b^2+ca)^2}{c+a}+\dfrac{(c^2+ab)^2}{a+b}\geq \dfrac{\frac{4}{9}(a+b+c)^4}{2(a+b+c)}=\dfrac{2}{9}(a+b+c)^3$$
Again by Cauchy-Schwarz inequality, we have
$$(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\geq 9$$

Finally, using this we get
$$S\geq \dfrac{2}{9}(a+b+c)^3\geq \dfrac{2(a+b+c)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\dfrac{2abc(a+b+c)^2}{ab+bc+ca}$$

Solution 3

Apply Cauchy-Schwarz inequality.
$$\left(\dfrac{(a^2+bc)^2}{b+c}+\dfrac{(b^2+ca)^2}{c+a}+\dfrac{(c^2+ab)^2}{a+b}\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\geq \left(\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\right)^2$$

Change variables as $x=b+c$, $y=c+a$, $z=a+b$.

We have

$$\dfrac{a^2+bc}{b+c}=\dfrac{a^2+bc+ab+ac}{b+c}-a=\dfrac{yz}{x}-a$$

Hence,
$$\dfrac{(a^2+bc)^2}{b+c}+\dfrac{(b^2+ca)^2}{c+a}+\dfrac{(c^2+ab)^2}{a+b}\geq \dfrac{\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}-\frac{x+y+z}{2}\right)^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Lemma. For all $x,y,z$ positive real numbers, we have
$$\dfrac{yz}{x}+\dfrac{zx}{y}+\dfrac{xy}{z}\geq x+y+z$$
Proof. This can be obtained by summing up the following inequalities resulting from Arithmetic-Geometric mean inequality.
\begin{align*}
 \dfrac{zx}{y}+\dfrac{xy}{z}\geq 2x\\
 \dfrac{xy}{z}+\dfrac{yz}{x}\geq 2y\\
 \dfrac{yz}{x}+\dfrac{zx}{y}\geq 2z\\
\end{align*}
Lemma. For all $a,b,c$ positive real numbers, we have
$$\dfrac{2}{b+c}+\dfrac{2}{c+a}+\dfrac{2}{a+b}\leq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$
Proof. This can be obtained by summing up the following inequalities resulting from Arithmetic-Harmonic mean inequality (or Cauchy-Schwarz).
\begin{align*}
 \dfrac{\frac{1}{b}+\frac{1}{c}}{2}\geq \dfrac{2}{b+c}\\
 \dfrac{\frac{1}{c}+\frac{1}{a}}{2}\geq \dfrac{2}{c+a}\\
 \dfrac{\frac{1}{a}+\frac{1}{b}}{2}\geq \dfrac{2}{a+b}\\
\end{align*}
With these two lemmas, we get
$$\dfrac{\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}-\frac{x+y+z}{2}\right)^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\geq \dfrac{\left(\frac{x+y+z}{2}\right)^2}{\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\dfrac{2(a+b+c)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\dfrac{2abc(a+b+c)^2}{ab+bc+ca}$$

Solution 4
By Titu's lemma the left-hand side is at least $\frac{(a^2 + b^2 + c^2 + ab + bc + ca)^2}{2(a+b+c)}$, so we want $(a^2 + b^2 + c^2 + ab + bc + ca)^2 \geq \frac{4abc(a+b+c)^3}{ab + bc + ca}$. However, $2(a^2 + b^2 + c^2 + ab + bc + ca) = (a+b)^2 + (b+c)^2 + (c+a)^2 \geq \frac{4(a+b+c)^2}{3}$ by Titu's lemma (or by the inequality between quadratic and arithmetic means, or by equivalence to $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$). Hence it remains to show $\frac{4(a+b+c)^4}{9} \geq \frac{4abc(a+b+c)^3}{ab + bc + ca}$, i.e. $(a+b+c)(ab + bc + ca) \geq 9abc$. The latter follows by multiplying the AM-GM-s $a+b+c \geq 3\sqrt[3]{abc}$ and $ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2}$.