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Question

Cho ba số thực dương a,b,c thỏa mãn a+b+c=3

Chứng minh $\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}\ge6$

Nguyễn Việt Lâm · Accepted Answer

Áp dụng BĐT

$a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\Rightarrow a^2+b^2+c^2\ge\dfrac{3^2}{3}=3$

Theo BĐT Bunhiacốpxki ta có:

$1.\sqrt{a^2+3}+1.\sqrt{b^2+3}+1.\sqrt{c^2+3}\ge\sqrt{\left(1+1+1\right)\left(a^2+b^2+c^2+9\right)}\ge\sqrt{3.\left(3+9\right)}=6$

Dấu "=" xảy ra khi $a=b=c=1$Câu trả lời: Áp dụng BĐT

$a^2+b^2+c^2\ge\dfrac{\left(a+b+c\right)^2}{3}\Rightarrow a^2+b^2+c^2\ge\dfrac{3^2}{3}=3$

Theo BĐT Bunhiacốpxki ta có:

$1.\sqrt{a^2+3}+1.\sqrt{b^2+3}+1.\sqrt{c^2+3}\ge\sqrt{\left(1+1+1\right)\left(a^2+b^2+c^2+9\right)}\ge\sqrt{3.\left(3+9\right)}=6$

Dấu "=" xảy ra khi $a=b=c=1$

tthnew · Answer

$VT=\sqrt{a^2+b^2+c^2+9+2\Sigma_{cyc}\sqrt{\left(a^2+1+1+1\right)\left(1+b^2+1+1\right)}}$

$\ge\sqrt{\frac{\left(a+b+c\right)^2}{3}+9+2\Sigma_{cyc}\left(a+b+1+1\right)}$

$=\sqrt{3+9+2\Sigma_{cyc}\left(2\left(a+b+c\right)+6\right)}=\sqrt{36}=6$

Is that true?Câu trả lời: $VT=\sqrt{a^2+b^2+c^2+9+2\Sigma_{cyc}\sqrt{\left(a^2+1+1+1\right)\left(1+b^2+1+1\right)}}$

$\ge\sqrt{\frac{\left(a+b+c\right)^2}{3}+9+2\Sigma_{cyc}\left(a+b+1+1\right)}$

$=\sqrt{3+9+2\Sigma_{cyc}\left(2\left(a+b+c\right)+6\right)}=\sqrt{36}=6$

Is that true?

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