Câu hỏi của N.T.M.D

Question

Cho a,b,c >0 thỏa mãn a+b+c$\le$$\frac{3}{2}$.Chứng minha,$\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$$\ge$6b,a+ b+ c+ $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$\ge$$\frac{15}{2}$

Yeutoanhoc · Accepted Answer

a)Áp dụng BĐT cosi-schwart:`A=1/a+1/b+1/c>=9/(a+b+c)`Mà `a+b+c<=3/2``=>A>=9:3/2=6`Dấu "=" `<=>a=b=c=1/2`b)Áp dụng BĐT cosi:`a+1/(4a)>=1``b+1/(4b)>=1``c+1/(4c)>=1``=>a+b+c+1/(4a)+1/(4b)+1/(4c)>=3`Ta có:`1/a+1/b+1/c>=6`(Ở câu a)`=>3/4(1/a+1/b+1/c)>=9/2``=>a+b+c+1/(a)+1/(b)+1/(c)>=3+9/2=15/2`Dấu "=" `<=>a=b=c=1/2`Câu trả lời: a)Áp dụng BĐT cosi-schwart:`A=1/a+1/b+1/c>=9/(a+b+c)`Mà `a+b+c<=3/2``=>A>=9:3/2=6`Dấu "=" `<=>a=b=c=1/2`b)Áp dụng BĐT cosi:`a+1/(4a)>=1``b+1/(4b)>=1``c+1/(4c)>=1``=>a+b+c+1/(4a)+1/(4b)+1/(4c)>=3`Ta có:`1/a+1/b+1/c>=6`(Ở câu a)`=>3/4(1/a+1/b+1/c)>=9/2``=>a+b+c+1/(a)+1/(b)+1/(c)>=3+9/2=15/2`Dấu "=" `<=>a=b=c=1/2`

Thành Trung Nguyễn Danh... · Answer

a)Áp dụng BĐT cosi-schwart:A=1a+1b+1c≥9a+b+cA=1a+1b+1c≥9a+b+cMà a+b+c≤32a+b+c≤32⇒A≥9:32=6⇒A≥9:32=6Dấu "=" ⇔a=b=c=12⇔a=b=c=12b)Áp dụng BĐT cosi:a+14a≥1a+14a≥1b+14b≥1b+14b≥1c+14c≥1c+14c≥1⇒a+b+c+14a+14b+14c≥3⇒a+b+c+14a+14b+14c≥3Ta có:1a+1b+1c≥61a+1b+1c≥6(Ở câu a)⇒34(1a+1b+1c)≥92⇒34(1a+1b+1c)≥92⇒a+b+c+1a+1b+1c≥3+92=152⇒a+b+c+1a+1b+1c≥3+92=152Dấu "=" ⇔a=b=c=12 Câu trả lời: a)Áp dụng BĐT cosi-schwart:A=1a+1b+1c≥9a+b+cA=1a+1b+1c≥9a+b+cMà a+b+c≤32a+b+c≤32⇒A≥9:32=6⇒A≥9:32=6Dấu "=" ⇔a=b=c=12⇔a=b=c=12b)Áp dụng BĐT cosi:a+14a≥1a+14a≥1b+14b≥1b+14b≥1c+14c≥1c+14c≥1⇒a+b+c+14a+14b+14c≥3⇒a+b+c+14a+14b+14c≥3Ta có:1a+1b+1c≥61a+1b+1c≥6(Ở câu a)⇒34(1a+1b+1c)≥92⇒34(1a+1b+1c)≥92⇒a+b+c+1a+1b+1c≥3+92=152⇒a+b+c+1a+1b+1c≥3+92=152Dấu "=" ⇔a=b=c=12

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)