2015 Algebra 2 - Problem 8

zschess wrote: Given positive reals $ a, b, c $ that satisfy $ a + b + c = 1 $, show that $$ \displaystyle \sum^{}_{cyc}\frac{a^3+bc}{a^2+bc}\ge 2 $$ We have \[\sum^{}_{cyc}\frac{a^3+bc}{a^2+bc}-2=\frac{abc\sum{(a^2-b^2)^2}}{2(a^2+bc)(ac+b^2)(ab+c^2)}\ge{0}\]

zschess wrote: Given positive reals $ a, b, c $ that satisfy $ a + b + c = 1 $, show that $$ \displaystyle \sum^{}_{cyc}\frac{a^3+bc}{a^2+bc}\ge 2 $$ Can prove by Cauchy-Schwarz

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zschess wrote: Given positive reals $ a, b, c $ that satisfy $ a + b + c = 1 $, show that $$ \displaystyle \sum^{}_{cyc}\frac{a^3+bc}{a^2+bc}\ge 2 $$ Given positive reals $ a, b, c $ that satisfy $ a + b + c = 1 $, show that \[{\frac {{a}^{2}+bc}{{a}^{3}+bc}}+{\frac {{b}^{2}+ca}{{b}^{3}+ca}}+{ \frac {{c}^{2}+ab}{{c}^{3}+ab}}\geq 3+{\frac {81}{2}}\,abc\]

$x,y,z>0,x+y+z=1$,prove that: \[{\frac {{x}^{2}+yz}{2\,{x}^{3}+7\,yz}}+{\frac {{y}^{2}+zx}{2\,{y}^{3}+ 7\,zx}}+{\frac {{z}^{2}+xy}{2\,{z}^{3}+7\,xy}}\geq {\frac {18}{23}} \] \[{\frac {{x}^{2}-5\,yz}{{x}^{3}+yz}}+{\frac {{y}^{2}-5\,zx}{{y}^{3}+zx} }+{\frac {{z}^{2}-5\,xy}{{z}^{3}+xy}}+9\geq 0\]

Let's expand the $LHS=\sum_{cyc} \frac{a^3+bc}{a^2+bc}=$ into this: $\sum_{cyc} \frac{a^3+abc+bc(b+c)}{a^2+bc}=$ which can be further simplified into this: $a+b+c+\sum_{cyc} \frac{bc(b+c)}{a^2+bc}.$ Therefore the above statement is equal to $\sum_{cyc} \frac{bc(b+c)}{a^2+bc} \ge a+b+c$. After we full expand the $LHS$ into this: $abc(a^4+b^4+c^4-a^2b^2-a^2c^2-b^2c^2) \ge 0$, we realize that this is trivial by AM-GM. Therefore we come to the conclusion that equality holds for $a=b=c=0$ for all non-negative integers.

Clearing the denominator, we get $3S(5,1,1) \geq 3S(3,3,1)$ Obviously this holds by Muirhead inequality

Hey , where am i going wrong? I am gettingit $ \geq 3$ !

I found a solution by Chebyshev: If we substract 1 from each term, we would get that we have to prove that \sum b+c/1+bc/a^2 is at most 1.But,from Chebyshev inequality we have that this sum is at most 2t/3,where t is \sum a^2/a^2+bc. On the other side,if we write bc as bc(a+b+c) then this turns to \sum b+c/1+bc/a^2 is at least 1.Again,by Chebyshev Inequality we get that this is at least 2t/3. Now we have the conclusion.

$\clubsuit \color{blue}{\textit{\textbf{Proof:}}}$ By homogenizing the inequality, we have to show that \[\sum_{cyc} \left(\frac{a^3+bc(a+b+c)}{a^2+bc}-a-b\right)\ge 0.\]This is equivalent to \[\sum_{cyc} \frac{a(b^2-c^2)}{c^2+ab}\ge 0,\]clearing the denominators yields \[a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\]which is true by AM-GM. $\quad \blacksquare$