Giúp tôi giải toán và làm văn

Mạnh hơn BĐT bài 2a:

Chứng minh rằng với mọi \(a,b>0;k\le8\) thì BĐT sau luôn đúng:\(\frac{k}{a^3+b^3}+\frac{1}{a^3}+\frac{1}{b^3}\ge\frac{16+4k}{\left(a+b\right)^3}\)

\(\Leftrightarrow\left(a-b\right)^2\left[\left(a^4+5a^3b+12a^2b^2+5ab^3+b^4\right)\left(a^2-ab+b^2\right)-3ka^3b^3\right]\ge0\) 

( BĐT cuối luôn đúng với mọi \(k\le8\) )

Thôi đành dồn về bậc dễ chịu hơn vậy :))
\(9=\frac{1}{a^3}+1+\frac{1}{a^3}+\frac{1}{b^3}+1+\frac{1}{b^3}+\frac{1}{c^3}+1+\frac{1}{c^3}\)

\(\ge\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le3\)

Đến đây ta có đánh giá bằng 2 cách như sau:

Cách 1:

Theo Bunhiacopski ta dễ có:

\(\left[2a+\left(b+c\right)\right]^2\ge4\cdot2a\left(b+c\right)\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{8a\left(b+c\right)}\)

\(\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{\left(b+c\right)^2}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{4bc}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8}\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\right]\)

Khi đó:

\(P\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8b^2}+\frac{1}{8c^2}+\frac{1}{4b^2}+\frac{1}{8a^2}+\frac{1}{8c^2}+\frac{1}{4c^2}+\frac{1}{8a^2}+\frac{1}{8b^2}\right]=\frac{3}{16}\)

Cách 2:

Áp dụng liên tiếp BĐT phụ dạng \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta dễ có rằng:

\(\frac{1}{\left(2a+b+c\right)^2}=\left(\frac{1}{2a+b+c}\right)^2=\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2=\frac{1}{16}\left[\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(a+c\right)^2}+\frac{2}{\left(a+b\right)\left(a+c\right)}\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(c+a\right)^2}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(b+c\right)\left(c+a\right)}+\frac{2}{\left(c+a\right)\left(a+b\right)}\)

\(\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

\(\le4\cdot\frac{1}{16}\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

\(=\frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\le\frac{1}{2}\cdot\left(3+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le3\)

\(\Rightarrow P\le\frac{3}{16}\)

Đẳng thức xảy ra tại a=b=c=1

1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.

2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)

\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)

\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)

\(\frac{a}{b+c+2a}=\frac{a}{\left(a+b\right)+\left(a+c\right)}\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}\)

Tương tự ta có: 

\(\frac{b}{a+c+2b}\le\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}\)

\(\frac{c}{a+b+2c}\le\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)

Cộng vế theo vế ta có: 

\(\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{c+b}+\frac{c}{a+c}+\frac{c}{b+c}\right)\)

\(=\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{a+c}{a+c}\right)=\frac{3}{4}\)

Dấu "=" xảy ra <=> a = b = c

Bổ sung bài 1:

BĐT được chứng minh

Đẳng thức xảy ra <=> a=b=c

Bài 1: diendantoanhoc.net

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành

\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)

\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)

Theo BĐT AM-GM và Cauchy-Schwarz ta có:

\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)

\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)

\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)

\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)

\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)

Bài 3 thì \(\le1\)

Bài 4 thì \(\ge\frac{3}{4}\) nhé

AD BĐT Bunhia dạng phân thức 

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu = khi a=b=c

=> Đpcm

Từ a²+b² >= 2ab => (a²+b²)\(\cdot\frac{1}{b}\ge2ab\cdot\frac{1}{b}\Rightarrow\frac{a^2}{b}+b\ge2a\left(1\right)\)

Tương tự \(\frac{b^2}{c}+c\ge2b\left(2\right);\frac{c^2}{a}+a\ge2c\left(3\right)\)

Từ (1) (2) (3) => đpcm

Sử dụng BĐT Cauchy-Schwarz ta có: \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{\left(a+b+c\right)^2}{ab+bc+ca}\)

Ta sẽ chứng minh \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}\ge\frac{9}{a+b+c}\Leftrightarrow\frac{9}{a+b+c}\le\frac{3}{ab+bc+ca}+2\)

Đặt a+b+c=t ta cần chứng minh \(\frac{6}{t^2-3}+2\ge\frac{9}{t}\Leftrightarrow\left(t+3\right)\left(t-3\right)^2\ge0\)

Dấu "=" xảy ra <=> a=b=c=1

Ok thanks, mặc dù ngay chỗ cuối đúng thì phải là (2t+3)(t-3)² >= 0
Nhưng hiểu rồi là OK :)

Ta có: 

\(15\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=10\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+2014\)

\(\le10\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2014\)

=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le\frac{2014}{5}\)

\(P=\frac{1}{\sqrt{5x^2+2xy+2yz}}+\frac{1}{\sqrt{5y^2+2yz+2zx}}+\frac{1}{\sqrt{5z^2+2zx+2xy}}\)

=> \(P\sqrt{\frac{2014}{135}}=\frac{1}{\sqrt{5x^2+2xy+2yz}.\sqrt{\frac{135}{2014}}}\)

\(+\frac{1}{\sqrt{5y^2+2yz+2zx}\sqrt{\frac{135}{2014}}}+\frac{1}{\sqrt{\frac{135}{2014}}\sqrt{5z^2+2zx+2xy}}\)

\(\le\frac{1}{2}\left(\frac{1}{5x^2+2xy+2yz}+\frac{2014}{135}+\frac{1}{5y^2+2yz+2zx}+\frac{2024}{135}+\frac{1}{5z^2+2yz+2zx}+\frac{2014}{135}\right)\)

\(\le\frac{1}{2}\left[\frac{1}{81}\left(\frac{5}{x^2}+\frac{2}{xy}+\frac{2}{yz}\right)+\frac{1}{81}\left(\frac{5}{y^2}+\frac{2}{yz}+\frac{2}{zx}\right)+\frac{1}{81}\left(\frac{5}{z^2}+\frac{2}{zx}+\frac{2}{xy}\right)+\frac{2014}{45}\right]\)

\(=\frac{5}{162}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2}{81}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{1007}{45}\)

\(\le\frac{5}{162}.\frac{2014}{5}+\frac{2}{81}.\frac{2014}{5}+\frac{1007}{45}=\frac{2014}{45}\)

=> \(P\le\frac{2014}{45}:\sqrt{\frac{2014}{135}}=3\sqrt{\frac{2014}{135}}\)

Dấu "=" xảy ra <=> x = y = z = \(\sqrt{\frac{15}{2014}}\)

Do 35 < x < 55

--> x = {36;37;38;...;52;53;54}

Mà những số chia hết cho 5 và 2 có đuôi là 0

--> x = 40;50

40 .Vì 40 chia hết cho cả  2 và5

40 bạn nhé

Áp dụng Cauchy Schwarz:

\(\frac{a^2}{a+1}+\frac{b^2}{b+1}\ge\frac{\left(a+b\right)^2}{a+b+2}=\frac{1}{3}\)

Đẳng thức xảy ra tại a=b=¹/₂

Ta có: 

\(\frac{a^2}{a+1}+\frac{b^2}{b+1}\)

\(=\frac{a^2}{a+1}+\frac{a+1}{9}+\frac{b^2}{b+1}+\frac{b+1}{9}-\frac{1}{3}\)

\(\ge2\sqrt{\frac{a^2}{a+1}.\frac{a+1}{9}}+2\sqrt{\frac{b^2}{b+1}.\frac{b+1}{9}}-\frac{1}{3}\)

\(=\frac{2}{3}a+\frac{2}{3}b-\frac{1}{3}=\frac{1}{3}\)

Dấu = xảy ra <=> a = b = 1/2

vãi lồn

!!!!!!!!!!!!!! WHAT? HOW TO?

😀 😁 ✍️ ☺️

Áp dụng BĐT Cosi ta có:

\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge2\sqrt{\frac{1}{\sqrt{ab}}}=\frac{2}{\sqrt{\sqrt{ab}}}\)

=> \(\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{\sqrt{ab}}{2}\)

<=> \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le2\cdot\frac{\sqrt{\sqrt{ab}}}{2}\)

<=> \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt{\sqrt{ab}}\)

Dấu "=" xảy ra <=> a=b

*Cô Chi check giúp em với ạ!*

Ta có: \(\sqrt{a}+\sqrt{b}\ge2\sqrt{\sqrt{ab}}\) ( cô - si cho hai số dương )

=> \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{2\sqrt{ab}}{2\sqrt{\sqrt{ab}}}=\sqrt{\sqrt{ab}}\)

Dấu "=" xảy ra <=> a = b.

Ta xét tống T của 3 số x(1-y);y(1-x);z(1-x)

Ta có T=x(1-y)+y(1-z)+z(1-x)=x+y+z-xy-xz-yz

Theo giả thiết xyz=(1-x)(1-y)(1-z)=1-(x+y+z-xy-xz-yz)-xyz

=> 2xyz=1-T => T=1-2xyz

Nhưng x²y²z² =[x(1-x)][y(1-y)][z(1-z)]\(\le\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{64}\)

=> xyz\(\le\)\(\frac{1}{8}\Rightarrow2xy\le\frac{1}{4}\)

Vậy \(T\ge1-\frac{1}{4}=\frac{3}{4}\)

Vậy \(T\ge\frac{3}{4}\)nên trong 3 số x(1-x), y(1-y), z(1-z) có ít nhất một trong 3 số đó \(\ge\frac{1}{4}\left(đpcm\right)\)

Bài 1:

Đặt a=x-1; b=y-1; c=z-1. Khi đó a;b;c\(\in\)[-1;1], a+b+c=0 và 

\(P=\left(a+1\right)^3+\left(b+1\right)^3+\left(c+1\right)^3-3abc\)

\(=a^3+b^3+c^3-3abc+3\left(a^2+b^2+c^2\right)+3\left(a+b+c\right)+3\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3\left(a^2+b^2+c^2\right)+3\left(a+b+c\right)+3\)

\(=3\left(a^2+b^2+c^2\right)+3\)

Ta có: \(0\le a^2+b^2+c^2\le2\)

Từ đây ta dễ thấy Min P=3 đạt được khi x=y=z=1

\(\left(a+b\right)\left(\frac{a}{b}+\frac{b}{a}\right)=\left(a+b\right)\left(\frac{a+b}{ab}\right)=\frac{\left(a+b\right)^2}{ab}=\frac{a^2+b^2+2ab}{ab}>=\frac{4ab}{ab}=4\)

Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

\(\sqrt{\frac{a}{a+bc}}=\frac{a}{\sqrt{a^2+abc}}=\frac{a}{\sqrt{a^2+ab+bc+ca}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

Tương tự \(\sqrt{\frac{b}{b+ca}}=\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}};\sqrt{\frac{c}{c+ab}}=\frac{c}{\left(c+a\right)\left(c+b\right)}\)

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

\(\le\frac{a}{2}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{b}{2}\left(\frac{1}{b+c}+\frac{1}{b+a}\right)+\frac{c}{2}\left(\frac{1}{c+a}+\frac{1}{c+b}\right)\)

\(=\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{c}{a+c}\right)\)

\(=\frac{3}{2}\)

Dấu "=" xảy ra tại \(a=b=c=3\)

Từ giả thiết ta có: (a+1)(b+1)(c+1) >=0 và (1-a)(1-b)(1-c) >=0

=> (a+1)(b+1)(c+1) +(1-a)(1-b)(1-c) >=0

Rút gọn ta có: -2((ab+bc+ca) =<2

Mặt khác (a+b+c)²=a²+b²+c²+2(ab+bc+ca)=0

=> a²+b²+c²=-2(ab+bc+ca)

=> a²+b²+c² =<2

Dấu "=" xảy ra <=> a=0; b=1; c=-1

co  mik đúng ik

A=x³+y³+2xy

<=> A=(x³+y³)+2xy

<=> A=(x+y)(x²-x+1)+2xy

mà x+y=2 => A=2(x²-x+1) +2xy

=> Min_A=2xy