# An inequality with $a^3+b^3+c^3=a^4+b^4+c^4$

Let $$a,b,c$$ are positive real numbers and $$a^3+b^3+c^3=a^4+b^4+c^4$$. Prove that: $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\ge1$$

My solve: So we have $$\frac{a^2}{a^3+ab^3+ac^3}+\frac{b^2}{b^3+bc^3+ba^3}+\frac{c^2}{c^3+ca^3+cb^3}\ge\frac{(a+b+c)^2}{a^3+b^3+c^3+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3}$$ Now because $$a^3+b^3+c^3=a^4+b^4+c^4$$ so $$a^3+b^3+c^3+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3$$$$=a^4+b^4+c^4+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3$$$$=(a+b+c)(a^3+b^3+c^3)$$ But I can't prove $$a+b+c=a^3+b^3+c^3$$. If you have an idea, then please let me know, thank you!

• Nov 4, 2021 at 10:54
• Hint:- $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)\ge a^{3}+b^{3}+c^{3}+3(a+b+c)$ Nov 4, 2021 at 10:55

Hint: For positive $$x$$, $$f(x)=(x-x^3)-2(x^3-x^4)=x(x-1)^2(2x+1)\geqslant 0$$. Hence $$f(a)+f(b)+f(c) = (a+b+c)-(a^3+b^3+c^3) \geqslant 0$$.