By Cauchy-Schwartz, we have $$\left(\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac {c^2+a^2}{a}\right)(b^5+bc^2+c^5+ca^2+a^5+ab^2)\geq (ab^2+bc+bc^2+ac+ca^2+ab)^2$$Now it is sufficient to show that $$(ab^2+bc+bc^2+ac+ca^2+ab)^2\geq 4(bc^2+ca^2+ab^2)(ab+bc+ca)$$This is true, since if we let $x=ab^2+bc^2+ca^2$ and $y=ab+bc+ca$, this is equivalent to $(x+y)^2\geq 4xy$, which is just AM-GM!

Com10atorics wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(ab+bc+ca).$$ By Cauchy-Schwarz we have \begin{align*} \left(\sum_{cyc}\frac{a^2}{b}\right)\left(\sum_{cyc}b^5\right)\geq\left(\sum_{cyc}ab^2\right)^2=\left(\sum_{cyc}ab^2\right)\left(\sum_{cyc}b^5\right)\\ \sum_{cyc}\frac{a^2}{b}\geq\sum_{cyc}ab^2 \end{align*}By AM-GM \[ \sum_{cyc}a+ab^2\geq\sum_{cyc}2\sqrt{a\cdot ab^2}=2(ab+bc+ca) \]Thus \[ \frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq ab^2+bc^2+ca^2+a+b+c\geq2(ab+bc+ca) \]

claim: $\sum{a} \ge \sum{a^3}$ from the condition using rearrangement $$\sum{a^3} \ge \sum {a.b^2}=\sum{a^5}$$by C.S $\sum{a^5}.\sum{a} \ge (\sum{a^3})^2$ so $$\sum{a} \ge \sum{a^3}$$$\blacksquare$ $$LHS = \sum{\frac{a^2}{b}}+ \sum{a } \ge \sum{\frac{a^2}{b}}+ \sum{b^3} \ge 2\sum{ab} =RHS$$and we win

Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^2+b^2+c^2)$$$$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^3+b^3+c^3)$$

Com10atorics wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(ab+bc+ca).$$ Solution of Zhangyanzong: $$\sum_{cyc}\frac{a^2+b^2}{b}=\sum_{cyc}\left(\frac{a^2}{b}+b^5\right)-\sum_{cyc} ab^2+\sum_{cyc}a\geq \sum_{cyc} a(b^2+1)\geq 2(ab+bc+ca).$$ sqing wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^3+b^3+c^3)$$ Solution of Zhangyanzong: $$\sum_{cyc}\frac{a^2+b^2}{b}=\sum_{cyc}\left(\frac{a^2}{b}+b^5\right)-\sum_{cyc} a^5+\sum_{cyc}a\geq \sum_{cyc} a(a^4+1)\geq 2(a^3+b^3+c^3).$$

sqing wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^2+b^2+c^2)$$ Any one?

sqing wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^3+b^3+c^3)$ just use rearrangement and the claim in my solution $$LHS = \sum \frac{a^2}{b}+\sum a \ge 2\sum{a} \ge 2\sum a^3$$

sqing wrote: sqing wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^2+b^2+c^2)$$ Any one? Nice one First of all by AM-GM we get: $$ b^3 + b^2 a + ba^2 \ge 3ab^2 $$$$ c^3 + c^2 b + cb^2 \ge 3bc^2 $$$$ a^3 + a^2 c + ac^2 \ge 3ca^2 $$By combining the above we get: $$ \sum a^3 + \sum_{sym} ab^2 \ge 3 \sum_{cyc} ab^2 $$$$ \implies \left(\sum a \right) \left( \sum a^2 \right) \ge 3 \sum_{cyc} ab^2 $$By Holder's inequality we have: $$ 3 \left(\sum a \right) \left(\sum a^5 \right) \ge \left(\sum a^2 \right)^3 $$Hence from the last two inequalities we get: $$ 3 \left(\sum a \right)^2 \left(\sum a^5 \right) \ge \left(\sum a \right) \left(\sum a^2 \right)^3 \ge 3 \left(\sum_{cyc} ab^2 \right) \left(\sum a^2 \right)^2 $$$$ \implies \sum a \ge \sum a^2 $$Now notice that by AM-GM we get: $$ \sum_{cyc} \frac{a^2+b^2}{b} \ge 2 \sum a \ge 2 \sum a^2 $$We are done. $\blacksquare$

By AM-GM, $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq \frac{2ab}{b}+\frac{2bc}{c}+\frac{2ca}{a}=2(a+b+c)\geq 2(a^3+b^3+c^3)$$

Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove that $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a+b+c)\geq 2(a^2+b^2+c^2)\geq 2(ab+bc+ca)$$$$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a+b+c)\geq 2(a^3+b^3+c^3)$$ Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove that $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^4+b^4+c^4)$$Solution : $$\sum_{cyc}\frac{a^2+b^2}{b}=\sum_{cyc}\left(\frac{a^2}{b}+b^5\right)-\sum_{cyc} a^5+\sum_{cyc}a\geq \sum_{cyc} a^3(a^2+1)\geq 2(a^4+b^4+c^4).$$

Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove that $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^n+b^n+c^n).$$Where $n=1,2,3,4.$ Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove or disprove $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^n+b^n+c^n).$$Where $n\in N^+.$

sqing wrote: Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove the inequality: $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^2+b^2+c^2)$$ By AM-GM $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a+b+c)$$$$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^3+b^3+c^3)$$$$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq a^3+a+b^3+b+c^3+c\geq 2(a^2+b^2+c^2)$$Solution of Zhangyanzong: $$\sum_{cyc}\frac{a^2+b^2}{b}\geq \sum_{cyc}\frac{2ab}{b}=\sum_{cyc}a+\sum_{cyc}a(a^4+1)-\sum_{cyc} ab^2$$$$\geq\sum_{cyc}a(a^2+1)+\frac{1}{3}\sum_{cyc}(a^3+b^3+b^3)-\sum_{cyc} ab^2\geq 2(a^2+b^2+c^2).$$

Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove that $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^5+b^5+c^5)$$Solution : By AM-GM, $$\sum_{cyc}\frac{a^2+b^2}{b}\geq \sum_{cyc}\left(\frac{a^2}{b}+b^5\right)+\frac{1}{3}\sum_{cyc}(a^3+b^3+b^3)-\sum_{cyc} ab^2\geq 2\sum_{cyc} ab^2= 2\sum_{cyc} a^5.$$Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove that $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq2(a+b+c)\geq 2(a^5+b^5+c^5)$$Solution : By AM-GM, $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq \frac{2ab}{b}+\frac{2bc}{c}+\frac{2ca}{a}=2(a+b+c)$$$$a+b+c=a+a^5+b+b^5+c+c^5-(a^5+b^5+c^5)\geq ab^2+bc^2+ca^2-(a^5+b^5+c^5)=a^5+b^5+c^5$$

Let $a,b$ and $c$ be positive real numbers such that $a^5+b^5+c^5=ab^2+bc^2+ca^2$. Prove or disprove $$\frac{a^2+b^2}{b}+\frac{b^2+c^2}{c}+\frac{c^2+a^2}{a}\geq 2(a^6+b^6+c^6).$$

Let $ a,b,c$ be positive real numbers such that $ 4abc = a + b + c + 1$. Prove that $$ \displaystyle{\frac {b^2 + c^2}{a} + \frac {c^2 + a^2}{b} + \frac {b^2 + a^2}{c}\geq 2(ab + bc + ca)}$$Let $ a,$ $ b$ and $ c$ are positive numbers such that $ 4abc = a + b + c + 1.$ Prove that $$ \frac {b^2 + c^2}{a} + \frac {c^2 + a^2}{b} + \frac {b^2 + a^2}{c}\geq 2(a^2 + b^2 + c^2)$$ for $a, b, c, > 0$ $$a+b+c+1=4abc$$is equivalent to $$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}=1.$$Let $a,b$ be positive real numbers such that $a+b+1=3ab.$ Prove that$$ \sqrt{a+1}+\sqrt{b+1}\leq \sqrt{2}( \sqrt{a}+\sqrt{b}).$$Let $ a,b>0 $ and $a+b+1=3ab .$ Prove that$$\frac{1}{a(b+1)}+\frac{1}{b(a+1)}\leq 1$$Let $x,y>0$ such that $ x+y+1=3xy$. Prove that $$\frac{3x}{y\left ( x+1 \right )}+\frac{3y}{x\left ( y+1 \right )}-\frac{1}{x^{2}}-\frac{1}{y^{2}}\leq 1$$https://artofproblemsolving.com/community/c6h1203942p6386705