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Prove that for $a,b,c>0$ $$\frac{a^2}{a^2+b^2+c^2-bc}+\frac{b^2}{a^2+b^2+c^2-ca}+\frac{c^2}{a^2+b^2+c^2-ab}\le\frac{3}{2}$$

Let $a,b,c>0$ and $a+b+c=3$ Prove that $$\frac{1}{\Big((a-b)^2+3bc+3ca\Big)^2}+\frac{1}{\Big((b-c)^2+3ca+3ab\Big)^2}+\frac{1}{\Big((c-a)^2+3ab+3bc\Big)^2}\ge \frac{1}{12}$$

Let $a,b,c>0$ Prove that $$\frac{a^3+b^3}{2c}+\frac{b^3+c^3}{2a}+\frac{c^3+a^3}{2b}\ge (a-b)^2+(b-c)^2+(c-a)^2+(a^2+b^2+c^2)$$

$\displaystyle \frac{1}{\Big((a-b)^2+3bc+3ca\Big)^2}+\frac{1}{\Big((b-c)^2+3ca+3ab\Big)^2}+\frac{1}{\Big((c-a)^2+3ab+3bc\Big)^2}$

$\displaystyle \ge \frac{1}{3} \Big( \frac{1}{(a-b)^2+3bc+3ca}+\frac{1}{(b-c)^2+3ca+3ab}+\frac{1}{(c-a)^2+3ab+3bc} \Big)^2$

$\displaystyle \ge \frac{1}{3} \Big( \frac{9}{2(a^2+b^2+c^2)+4(ab+bc+ca)}\Big)^2 = \frac{1}{3} \cdot \Big( \frac{9}{2\cdot 3^2} \Big)^2 = \frac{1}{12}$

$\displaystyle \frac{a^3+b^3}{2c}+\frac{b^3+c^3}{2a}+\frac{c^3+a^3}{2b} - a^2-b^2-c^2$

$\displaystyle =\frac{(a^3-c^3)+(b^3-c^3)}{2c}+\frac{(b^3-a^3)+(c^3-a^3)}{2a}+\frac{(c^3-b^3)+(a^3-b^3)}{2b}$

$\displaystyle = (b^3-a^3)(\frac{1}{2a}-\frac{1}{2b})+(c^3-b^3)(\frac{1}{2b}-\frac{1}{2c})+(a^3-c^3)(\frac{1}{2c}-\frac{1}{2a})$

$\displaystyle = (b-a)^2(\frac{a^2+ab+b^2}{2ab})+ (c-b)^2(\frac{b^2+bc+c^2}{2bc})+ (a-c)^2(\frac{c^2+ca+a^2}{2ca})$

$\displaystyle \ge \frac{3}{2} ((a-b)^2+(b-c)^2+(c-a)^2)$

ก็จะได้ว่ามากกว่า $(a-b)^2+(b-c)^2+(c-a)^2$ ด้วยครับ

$\displaystyle \frac{a^2}{a^2+b^2+c^2-bc}+\frac{b^2}{a^2+b^2+c^2-ca}+\frac{c^2}{a^2+b^2+c^2-ab}$

$\displaystyle \le \frac{a^2}{a^2+\frac{1}{2}(b^2+c^2)}+\frac{b^2}{b^2+\frac{1}{2}(c^2+a^2)}+\frac{c^2}{c^2+\frac{1}{2}(a^2+b^2)}$

$\displaystyle = \frac{2a^2}{(a^2+b^2)+(a^2+c^2)}+\frac{2b^2}{(b^2+c^2)+(b^2+a^2)}+\frac{2c^2}{(c^2+a^2)+(c^2+b^2)}$

$\displaystyle \le \frac{1}{2}\Big( \frac{a^2}{a^2+b^2}+\frac{a^2}{a^2+c^2}\Big)+\frac{1}{2}\Big(\frac{b^2}{b^2+a^2}+\frac{b^2}{b^2+c^2}\Big)+\frac{1}{2}\Big(\frac{ c^2}{c^2+a^2}+\frac{c^2}{c^2+b^2}\Big)$

$\displaystyle = \frac{3}{2}$

ครบทุกข้อแล้วครับ :great::great:

เก่งมากๆเลยครับ :great: