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Let $a,b,c$ be the length of sides of triangle such that $a+b+c=2$. Prove that $$21(a^2+b^2+c^2)\ge 20 +9(a^3+b^3+c^3)$$

It was in my exam. It can be solved easy by BW but it takes alot of time to expand. I tried to use Schur, AM-GM but all failed.Help me solve it without bw.

Michael Rozenberg
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user774564
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3 Answers3

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Let $a=\frac{2}{3}x$, $b=\frac{2}{3}y$ and $c=\frac{2}{3}z$.

Thus, $x$, $y$ and $z$ are also sides-lengths of a triangle, $x+y+z=3$ and we need to prove that: $$7(x^2+y^2+z^2)\geq15+2(x^3+y^3+z^3)$$ or $$\sum_{cyc}(2x^3-7x^2+5)\leq0$$ or $$\sum_{cyc}(x-1)(2x^2-5x-5)\leq0$$ or $$\sum_{cyc}((x-1)(2x^2-5x-5)+8(x-1))\leq0$$ or $$\sum_{cyc}(x-1)^2(3-2x)\geq0$$ or $$\sum_{cyc}(x-1)^2(y+z-x)\geq0$$ and we are done!

Michael Rozenberg
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Your inequality equivalent to $$21(a^2+b^2+c^2) \cdot \frac{a+b+c}{2}\ge \frac52(a+b+c)^3 +9(a^3+b^3+c^3).$$ Using Ravi substitution for $a = x+y,\,b=y+z,\,c=z+x$ with $x,\,y,\,z > 0$ inequality become $$4(x^3+y^3+z^3)+6xyz \geqslant 3[xy(x+y)+yz(y+z)+zx(z+x)]. \quad (1)$$ From Schur inequality $$x^3+y^3+z^3 +3xyz \geqslant xy(x+y)+yz(y+z)+zx(z+x),$$ we need to prove $$2(x^3+y^3+z^3) \geqslant xy(x+y)+yz(y+z)+zx(z+x).$$ Which is true because $x^3+y^3 \geqslant xy(x+y).$

Note. The sum of squares of $(1)$ $$x(y+z-2x)^2+y(z+x-2y)^2+z(x+y-2z)^2 \geqslant 0.$$

Nguyenhuyen_AG
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We write inequality as $$21(a^2+b^2+c^2) \cdot \frac{a+b+c}{2} \geqslant \frac52(a+b+c)^3 +9(a^3+b^3+c^3),$$ or $$42s(s^2-4Rr-r^2) \geqslant 20s^3+18s(s^2-6Rr-3r^2),$$ or $$4s(s^2+3r^2-15Rr) \geqslant 0.$$ But $$s^2+3r^2-15Rr = r(R-2r)+(s^2-16Rr+5r^2) \geqslant 0.$$ Which is true by Euler's inequality and Gerretsen's inequality.

Nguyenhuyen_AG
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