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Problem 1. For all positive real numbers , prove that
![\[
\frac{(a^2 + bc)^2}{b + c} + \frac{(b^2 + ca)^2}{c + a} + \frac{(c^2 + ab)^2}{a + b} \geq \frac{2abc(a + b + c)^2}{ab + bc + ca}.
\]](//latex.artofproblemsolving.com/e/7/0/e70d68c260be93ad221f67dbaa3adfd2b121a312.png)
Solution 1Solution 2Solution 3Solution 4
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 . 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:
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\frac{a + b}{2} \geq \sqrt{ab}
\]](//latex.artofproblemsolving.com/1/7/2/172033d904b47abbb0b3a8e25946ede24ca24599.png)
In this case, we apply the AM-GM inequality as follows:
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\sum_{\text{cyc}} \frac{(a^2 + bc)^2}{b+c} \geq \sum_{\text{cyc}} \frac{4a}{b+c}
\]](//latex.artofproblemsolving.com/1/c/c/1ccba803b632d8af292f2d04fc43d0196bce28e4.png)
So, by applying AM-GM, we get the first inequality:
Step 3: Second Inequality
In this step, we simplify the expression on the right-hand side further. The previously obtained sum:
![\[
\sum_{\text{cyc}} \frac{4a}{b+c}
\]](//latex.artofproblemsolving.com/e/d/0/ed04d6c38b2ff60133638289564ec2346590f5e7.png)
is rewritten as:
![\[
\sum_{\text{cyc}} \frac{4a^2}{ab+ac}
\]](//latex.artofproblemsolving.com/8/0/3/80326831675080475af9bd769a65db2ff15deee0.png)
Here, the terms are reorganized and simplified.
Step 4: Conclusion
Finally, we arrive at the inequality:
![\[
\sum_{\text{cyc}} \frac{4a^2}{ab+ac} \geq \frac{2(a + b + c)^2}{ab + bc + ca}
\]](//latex.artofproblemsolving.com/6/e/3/6e3b925f22cd0a13a62bdf84559dd11461a6456b.png)
This inequality can then be written as:
![\[
\frac{2(a + b + c)^2}{ab + bc + ca} = \frac{2abc(a + b + c)^2}{ab + bc + ca}
\]](//latex.artofproblemsolving.com/a/0/c/a0cc3e750aa4855cf16019464a0d359983ef6159.png)
Thus, the two sides are equal, and the desired result is achieved.
Solution 4
By Titu's lemma the left-hand side is at least , so we want . However,  by Titu's lemma (or by the inequality between quadratic and arithmetic means, or by equivalence to ). Hence it remains to show , i.e. . The latter follows by multiplying the AM-GM-s ![$a+b+c \geq 3\sqrt[3]{abc}$](//latex.artofproblemsolving.com/d/9/b/d9b5cff6e5b76539be1c057166c83b2d9969e9ed.png) and .
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