sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ Proof of Zhangyanzong: $$\sum_{cyc}\frac{b^2+1}{a+bc}=\sum_{cyc}\frac{b(b+ca)}{a+bc}\geq 3\iff \sum_{cyc}\frac{a^2-b^2}{a+bc}\leq \sum_{cyc}a-3.$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^3-b^3}{a+bc}+\frac{b^3-c^3}{b+ca}+\frac{c^3-a^3}{c+ab}\leq a+b+c-3.$$$$\frac{a^2-b^2}{1+bc}+\frac{b^2-c^2}{1+ca}+\frac{c^2-a^2}{1+ab}\geq a+b+c-3.$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ Just notice that \begin{align*} \sum_{\mathrm{cyc}} \left(a - \frac{a^2-b^2}{a+bc}\right) &= \sum_{\mathrm{cyc}} \left(a - \frac{a^2-b^2}{a+\frac{1}{a}}\right) \\ &= \sum_{\mathrm{cyc}} \frac{(a^2+1)-(a^2-b^2)}{a+\frac{1}{a}} \\ &= \sum_{\mathrm{cyc}} \frac{a(b^2+1)}{a^2+1} \\ &\geqslant 3\sqrt[3]{\frac{a(b^2+1)}{a^2+1}\cdot \frac{b(c^2+1)}{b^2+1}\cdot \frac{c(a^2+1)}{c^2+1}} \\ &= 3 \end{align*}

sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$

Attachments:

sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ Let $a_1,a_2,\cdots,a_n$ $(n\geq 3)$ be positive reals such that $a_1a_2\cdots a_n= 1$. Prove that $$\frac{a^2_1-a^2_2}{a_1+\frac{1}{a_1}}+\frac{a^2_2-a^2_3}{a_2+\frac{1}{a_2}}+\cdots+\frac{a^2_{n-1}-a^2_n}{a_{n-1}+\frac{1}{a_{n-1}}}+\frac{a^2_n-a^2_1}{a_n+\frac{1}{a_n}}\leq a_1+a_2+\cdots+a_n-n.$$

sqing wrote: Let $a_1,a_2,\cdots,a_n$ $(n\geq 3)$ be positive reals such that $a_1a_2\cdots a_n= 1$. Prove that $$\frac{a^2_1-a^2_2}{a_1+\frac{1}{a_1}}+\frac{a^2_2-a^2_3}{a_2+\frac{1}{a_2}}+\cdots+\frac{a^2_{n-1}-a^2_n}{a_{n-1}+\frac{1}{a_{n-1}}}+\frac{a^2_n-a^2_1}{a_n+\frac{1}{a_n}}\leq a_1+a_2+\cdots+a_n-n.$$ Solution. By the inequality AM-GM and the assumption $a_1a_2\cdots a_n= 1$, we deduce that \begin{align*}&a_1-\frac{a^2_1-a^2_2}{a_1+\frac{1}{a_1}}+a_2-\frac{a^2_2-a^2_3}{a_2+\frac{1}{a_2}}+\cdots+a_{n-1}-\frac{a^2_{n-1}-a^2_n}{a_{n-1}+\frac{1}{a_{n-1}}}+a_n-\frac{a^2_n-a^2_1}{a_n+\frac{1}{a_n}}\\ =&a_1\cdot\frac{1+a^2_2}{1+a_1^2}+a_2\cdot\frac{1+a^2_3}{1+a_2^2}+\cdots+a_{n-1}\cdot\frac{1+a^2_n}{1+a_{n-1}^2}+a_n\cdot\frac{1+a^2_1}{1+a_n^2}\\ \ge&n\sqrt[n]{a_1a_2\cdots a_n\cdot1}=n, \end{align*}which gives the desired inequality. $\blacksquare$

sqing wrote: sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^3-b^3}{a+bc}+\frac{b^3-c^3}{b+ca}+\frac{c^3-a^3}{c+ab}\leq a+b+c -3$$ We know that, $$ \sum_{cyc} a^2 \geq 3 \implies \sum_{cyc} b^3-a^3+a^2 \geq 3 \implies \sum_{cyc} \frac {b^3-a^3+a^2+1}{2} \geq 3$$ This can re-written as $$-(\sum_{cyc} \frac {b^3-a^3+a^2+1}{2} )\leq -3 \implies a+b+c-3 \geq \sum_{cyc} a - (\frac {b^3-a^3+a^2+1}{2} )$$ Now coming back to the question: $$(\sum_{cyc} a) -3 \geq \sum_{cyc} \frac {a^3-b^3}{a+bc} \geq \sum_{cyc} \frac {a (a+bc) -1-a^2+a^3-b^3}{a+bc} \geq \sum_{cyc} a - \frac {b^3-a^3+a^2+1}{a+\tfrac {1}{a}} \geq \sum_{cyc} a - \frac {b^3-a^3+a^2+1}{2} $$Which we have already proved above!!

BEAUTIFUL SOLUTION

sqing wrote: Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$ We know that $$\sum_{cyc} \frac {a (b^2+1)}{a^2+1} \geq 3\sqrt [3]{abc} \geq 3$$ This can be re-written as: $$(\sum_{cyc} a) -3 \geq \sum_{cyc} (a-\frac {a (b^2+1)}{a^2+1}) $$which is a proved inequality as for now Now back to the main problem: $$(\sum_{cyc} a) -3 \geq \sum_{cyc} \frac {a^2-b^2}{a+bc} = \sum_{cyc} \frac {a^3-ab^2}{a^2+1} = \sum_{cyc} \frac {a (a^2+1) -a -ab^2}{a^2+1} = \sum_{cyc} (a-\frac {a (b^2+1)}{a^2+1}) $$

$$abc=1$$$$\sum_{cyc}(a-\frac{a^2-b^2}{a+bc})=\sum_{cyc}(\frac{b(b+ac)}{a+bc})\ge 3$$by AM-GM very easy.

Notice that we can rearrange the inequality so that we have $\sum_{cyc}a-\frac{a^2-b^2}{a+b}\ge3$ Furthermore, by our conditions we can rewrite the inequality as $\sum_{cyc}a-\frac{a^2-b^2}{a+\frac{1}{a}}=\sum_{cyc}a-\frac{a^3+ab^2}{a^2+1}=\sum_{cyc}\frac{a(1+b^2)}{a^2+1}$ Thus the inequality boils down to $\sum_{cyc}\frac{a(1+b^2)}{a^2+1}\ge3$. However notice that $\sum_{cyc}\frac{a(1+b^2)}{a^2+1}\overset{\text{AM-GM}}{\ge}3\sqrt[3]{\frac{a(1+b^2)}{1+a^2}\cdot\frac{b(1+c^2)}{1+b^2}\cdot\frac{c(1+a^2)}{1+c^2}}=3\sqrt[3]{abc}=3$ $\blacksquare$.