Cho$a,b,c>0$ thỏa: $a+b+c=1$
Chứng minh rằng: $\sum \frac{\sqrt{a}}{1-a}\geqslant \frac{3\sqrt{3}}{2}$
Cho$a,b,c>0$ thỏa: $a+b+c=1$
Chứng minh rằng: $\sum \frac{\sqrt{a}}{1-a}\geqslant \frac{3\sqrt{3}}{2}$
To the extent math refers to reality, we are not certain;
to the extent we are certain, math does not refer to reality.~~Albert Einstein
Cm như sau
Ta đi cm $\frac{\sqrt{a}}{1-a}\geq \frac{3\sqrt{3}}{2}.a\Leftrightarrow \frac{1}{1-a}\geq \frac{3\sqrt{3}}{2}.\sqrt{a}\Leftrightarrow \sqrt{a}(1-a)\leq \frac{2}{3\sqrt{3}}\Leftrightarrow a(1-a)^{2}\leq \frac{4}{27}\Leftrightarrow \frac{1}{2}2a(1-a)(1-a)\leq \frac{4}{27}$
Mà $\frac{1}{2}2a(1-a)(1-a)\leq \frac{1}{2}(\frac{2a+1-a+1-a}{3})^{3}=\frac{4}{27}$
cmtt cộng theo vế ta đc đpcm
๖ۣۜI will try my best ๖ۣۜ
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