Cho $a,b,c\ge 0$ thỏa mãn $a+b+c=1$. Chứng minh: $\\ \sum\limits_{\text{cyc}}^{} \dfrac{a}{2a^3+1}\ge \dfrac{1}{3}$

Với $\dfrac{a}{2{{a}^{3}}+1}+\dfrac{b}{2{{b}^{3}}+1}+\dfrac{c}{2{{c}^{3}}+1}\ge \dfrac{1}{3}$

$\Leftrightarrow -\dfrac{a}{2{{a}^{3}}+1}-\dfrac{b}{2{{b}^{3}}+1}-\dfrac{c}{2{{c}^{3}}+1}\le -\dfrac{1}{3}$

$\Leftrightarrow \left( a-\dfrac{a}{2{{a}^{3}}+1} \right)+\left( b-\dfrac{b}{2{{b}^{3}}+1} \right)+\left( c-\dfrac{c}{2{{c}^{3}}+1} \right)\le \left( a+b+c \right)-\dfrac{1}{3}$

$\Leftrightarrow \dfrac{2{{a}^{4}}}{2{{a}^{3}}+1}+\dfrac{2{{b}^{4}}}{2{{b}^{3}}+1}+\dfrac{2{{c}^{4}}}{2{{c}^{3}}+1}\le \dfrac{2}{3}$

Như vậy, chuyển điều phải chứng minh trở thành:

$\dfrac{2{{a}^{4}}}{2{{a}^{3}}+1}+\dfrac{2{{b}^{4}}}{2{{b}^{3}}+1}+\dfrac{2{{c}^{4}}}{2{{c}^{3}}+1}\le \dfrac{2}{3}$

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Ta có ${{a}^{3}}+{{a}^{3}}+1\ge 3{{a}^{2}}$

$\Rightarrow 2{{a}^{3}}+1\ge 3{{a}^{2}}$

$\Rightarrow \dfrac{2{{a}^{2}}}{2{{a}^{3}}+1}\le \dfrac{2}{3}$

$\Rightarrow \dfrac{2{{a}^{4}}}{2{{a}^{3}}+1}\le \dfrac{2}{3}{{a}^{2}}$

Tương tự: $\begin{cases}\dfrac{2b^4}{2b^3+1}\le\dfrac{2}{3}b^2\\\dfrac{2c^4}{2c^3+1}\le\dfrac{2}{3}c^2\end{cases}$

$\Rightarrow \dfrac{2{{a}^{4}}}{2{{a}^{3}}+1}+\dfrac{2{{b}^{4}}}{2{{b}^{3}}+1}+\dfrac{2{{c}^{4}}}{2{{c}^{3}}+1}\le \dfrac{2}{3}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\,\,\,\,\,\left( 1 \right)$

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Do $a,b,c\ge 0$ và $a+b+c=1$

$\Rightarrow 0\le a,b,c\le 1$

$\Rightarrow {{a}^{2}}\le a$ và ${{b}^{2}}\le b$ và ${{c}^{2}}\le c$

$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\le a+b+c$

$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\le 1\,\,\,\,\,\left( 2 \right)$

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Từ $\left( 1 \right)$ và $\left( 2 \right)\Rightarrow \dfrac{2{{a}^{4}}}{2{{a}^{3}}+1}+\dfrac{2{{b}^{4}}}{2{{b}^{3}}+1}+\dfrac{2{{c}^{4}}}{2{{c}^{3}}+1}\le \dfrac{2}{3}$ (ĐPCM)

Dấu “=” xảy ra khi $\left( a;b;c \right)=\left( 0;0;1 \right)$ và các hoán vị