Bài 39: Cho $a,b,c$ là các số thực dương. Chứng minh rằng:
$\sqrt{a^{4}+\dfrac{b^{4}}{2}+\dfrac{c^{4}}{2}}+\sqrt{b^{4}+\dfrac{c^{4}}{2}+\dfrac{a^{4}}{2}}+\sqrt{c^{4}+\dfrac{a^{4}}{2}+\dfrac{b^{4}}{2}}\geq \sqrt{a^{4}+b^{3}c}+\sqrt{b^{4}+c^{3}a}+\sqrt{c^{4}+a^{3}b}$
P/s: Mọi người hãy post lên cả những ý tưởng nhé!
$\sum_{cyc}\sqrt{a^{4}+\dfrac{b^{4}}{2}+\dfrac{c^{4}}{2}}\geq\sum_{cyc}\sqrt{a^{4}+b^{3}c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(a^{4}-a^{3}b+2\sqrt{\left(a^{4}+\dfrac{b^{4}+c^{4}}{2}\right)\left(b^{4}+\dfrac{a^{4}+c^{4}}{2}\right)}\right)\geq2\sum_{cyc}\sqrt{(a^{4}+b^{3}c)(b^{4}+c^{3}a)}$.
Vì $2\sum_{cyc}(a^{4}+a^{3}b)\geq2\sum_{cyc}\sqrt{(a^{4}+b^{3}c)(b^{4}+c^{3}a)}$. Do đó cần cm
$2\sum_{cyc}\sqrt{\left(a^{4}+\dfrac{b^{4}+c^{4}}{2}\right)\left(b^{4}+\dfrac{a^{4}+c^{4}}{2}\right)}\geq \sum_{cyc}(a^{4}+3a^{3}b)$
Vì $2\sum_{cyc}\sqrt{\left(a^{4}+\dfrac{b^{4}+c^{4}}{2}\right)\left(b^{4}+\dfrac{a^{4}+c^{4}}{2}\right)}\geq2\sum_{cyc}\left(a^{2}\cdot\sqrt{\dfrac{a^{4}+c^{4}}{2}}+b^{2}\cdot\sqrt{\dfrac{b^{4}+c^{4}}{2}}\right)\geq$
$\geq2\sum_{cyc}\left(a^{2}\cdot\dfrac{a^{2}+c^{2}}{2}+b^{2}\cdot\dfrac{b^{2}+c^{2}}{2}\right)=2\sum_{cyc}(a^{4}+a^{2}b^{2})\geq\sum_{cyc}(a^{4}+3a^{3}b).$