Cho a, b, c thỏa mãn: abc=1 và \(a^3>36\). CMR: \(\dfrac{a^2}{2}+b^2+c^2>ab+bc+ca\)
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NT
0
HT
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
HN
0
24 tháng 6 2021
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
AT
24 tháng 6 2021
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}=\dfrac{\sqrt{ab+2c^2}}{\sqrt{a^2+b^2+ab}}=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+2c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+a^2+b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta cần chứng minh: \(\dfrac{a^2}{2}+b^2+c^2>ab+bc+ca\Leftrightarrow\dfrac{a^2}{2}+b^2+c^2-ab-bc-ca>0\Leftrightarrow\dfrac{a^2}{4}+b^2+c^2+ab+ca+2bc-3bc+\dfrac{a^2}{4}>0\) \(\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2}{12}+\dfrac{a^2}{6}-3bc>0\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) Mà \(a^3>36;abc=1\Rightarrow a^3>36abc\Rightarrow a^2>36bc\)
\(\Rightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) luôn đúng
Này Nguyễn Trọng Chiến, mk ko hiểu cái chỗ tách ra thành: \(\dfrac{a^2}{4}+b^2+c^2+ab+ca+2bc-3bc+\dfrac{a^2}{4}>0\). Sao bn tách đc vậy??