Question

Cho các số dương a,b,c thỏa mãn a+b+c=4

CMR: \(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}>4\)

Answer 1

\(VT=\sqrt{4-a}+\sqrt{4-b}+\sqrt{4-c}\)

Ta có BĐT phụ \(\sqrt{4-a}>-\dfrac{1}{2}a+2\)

\(\Leftrightarrow-\dfrac{1}{4}a\left(a-4\right)>0\forall0< a< 4\) (đúng)

Tương tự cho 2 BĐT còn lại cũng có:

\(\sqrt{4-b}>-\dfrac{1}{2}b+2;\sqrt{4-c}>-\dfrac{1}{2}c+2\)

Cộng theo vế 3 BĐT trên ta có:

\(VT>-\dfrac{1}{2}\left(a+b+c\right)+6=4=VP\)

Answer 2

ta có:

\(2VT=\sqrt{4\left(a+b\right)}+\sqrt{4\left(a+c\right)}+\sqrt{4\left(c+a\right)}\)

\(=\sqrt{\left(a+b+c\right)\left(a+b\right)}+\sqrt{\left(a+b+c\right)\left(b+c\right)}+\sqrt{\left(a+b+c\right)\left(c+a\right)}\)

\(>\sqrt{\left(a+b\right)^2}+\sqrt{\left(b+c\right)^2}+\sqrt{\left(c+a\right)^2}=2\left(a+b+c\right)=8\)

\(\Rightarrow VT>4\)(đpcm)

Answer 3

Answer 4

Từ [TEX]\frac{a}{c}= \frac{b}{d}
ightarrow \frac{4a}{6c}= \frac{4b}{6d}
ightarrow \frac{4a^2}{6ac}= \frac{4b^2}{6bd}[/TEX]

[TEX]
ightarrow \frac{4a^2}{4b^2}{6ac}{6bd}= \frac{4a^2-6ac}{4b^2-6bd}= \frac{4a^2+6ac}{4b^2+6bd}[/TEX]

Do đó [TEX]\frac{4a^2-6ac}{4b^2-6bd}= \frac{4a^2+6ac}{4b^2+6bd}
ightarrow ( 4a^2 - 6ac)(4b^2 + 6bd) =( 4a^2 +6ac)(4b^2 - 6bd)[/TEX].

Ta có đpcm.