Doubt:regarding proof $\sum_{cyc}a^{2/3}b\le 3$

Question

i had a doubt regarding the proof given in my textbook

prove $$\sum_{cyc}a^{2/3}b\le 3$$ if $a,b,c>0$ and $a+b+c=3$ , the proof given is as follows

$$3\sum_{cyc}a\ge \sum_{cyc}a+2\sum_{cyc}ab\ge \sum_{cyc}(a+ac+ac)\ge 3\sum_{cyc}ac^{2/3}$$

what i dont understand is how they used $$\sum_{cyc}a\ge \sum_{cyc}ab$$.

Answer 1

We have known inequality $(a+b+c)^2 \geqslant 3(ab+bc+ca).$ So, if $a+b+c=3$ then $$a +b+c \geqslant ab+bc+ca$$

Answer 2

Another way.

We need to prove that: $$a^3b^2+b^3c^2+c^3a^2\leq3$$ for non-negative $a$, $b$ and $c$ such that $a^2+b^2+c^2=3.$

Indeed, let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.

Thus, by Rearrangement and AM-GM we obtain: $$a^3b^2+b^3c^2+c^3a^2=a\cdot a^2b^2+b\cdot b^2c^2+c\cdot c^2a^2\leq x\cdot x^2y^2+y\cdot x^2z^2+z\cdot y^2z^2=$$ $$=y(x^3y+x^2z^2+yz^3)=y\left(x^2\left(xy+\frac{z^2}{2}\right)+z^2\left(yz+\frac{x^2}{2}\right)\right)\leq$$ $$\leq y\left(x^2\left(\frac{x^2+y^2}{2}+\frac{z^2}{2}\right)+z^2\left(\frac{y^2+z^2}{2}+\frac{x^2}{2}\right)\right)=\frac{3}{2}y(3-y^2)\leq3.$$