Question by Lê Quốc Trần Anh - Discuss with MathYouLike

Lê Quốc Trần Anh Coordinator

02/04/2018 at 11:28

If a,b,c >0 and $a^{2}+b^{2}+c^{2}+abc=4$
Prove that : $a+b+c\geq \sqrt{a}+\sqrt{b}+\sqrt{c}$

List of answers

¤« 03/04/2018 at 13:20

Wrong post

Because if a,b,c > 0 so

⎧⎩⎨⎪⎪a≥a−−√b≥b√c≥c√⇒a+b+c≥a−−√+b√+c√
FC Alan Walker 03/04/2018 at 22:25

Dao Trong Luan is wrong.

When \(0< a,b,c< 1\), \(\sqrt{a}>a;\sqrt{b}>b;\sqrt{c}>c\)

Ex: \(\sqrt{0.36}=0.6>0.36\)
Dao Trong Luan Coordinator 02/04/2018 at 22:07

Wrong post

Because if a,b,c > 0 so

\(\left\{{}\begin{matrix}a\ge\sqrt{a}\\b\ge\sqrt{b}\\c\ge\sqrt{c}\end{matrix}\right.\Rightarrow a+b+c\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)

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