Áp dụng BĐT cô si với hai số không âm, Ta có: 

\(\left(a+b+c\right)^2=1\ge4a\left(b+c\right)\)

\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\)

Mà \(\left(b+c\right)^2\ge4bc\forall b,c\ge0\)

\(\Rightarrow b+c\ge16abc\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}a+b+c=1\\b=c\\a=b+c\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)

Áp dụng BĐT Cô si với 2 số dương ta có: 

\(\frac{a}{b}+\frac{b}{a}\ge2,\frac{b}{c}+\frac{c}{b}\ge2,\frac{c}{a}+\frac{a}{c}\ge2\)

\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)(đúng) 

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(do a+b+c=1)

3)

1.

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)

Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)

\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)

\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)

\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)

\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)

\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)

\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)

\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)

2.

Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)

Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)

\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)

Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)

\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)

\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)

(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị

Đặt  \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)

=> \(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

Đặt A < (1/40+.....+1/40)+(1/60+1/60+...+1/60)

=>A<1/2+1/3=5/6<3/2

lớn hơn 11/15 cũng tương tự thôi bạn tự làm sẽ thú vị hơn đấy

k minh nha

Thank you

√

bài 1b

+)Nếu n chẵn ,ta có \(n^4⋮2,4^n⋮2\Rightarrow n^4+4^n⋮2\)

mà \(n^4+4^n>2\)Do đó \(n^4+4^n\)là hợp số

+)nếu n lẻ đặt \(n=2k+1\left(k\in N\right)\)

Ta có \(n^4+4^n=n^4+4^{2k}.4=\left(n^2+2.4k\right)^2-2n^2.2.4^k\)

\(=\left(n^2+2^{2k+1}\right)^2-\left(2.n.2^k\right)^2\)

\(=\left(n^2+2^{2k+1}+2n.2^k\right)\left(n^2+2^{2k+1}-2n.2^k\right)\)

\(=\left(\left(n+2^k\right)^2+2^{2k}\right)\left(\left(n-2^k\right)^2+2^{2k}\right)\)

là hợp số,vì mỗi thừa số đều lớn hơn hoặc bằng 2

(nhớ k nhé)

Bài 2a)

Nhân 2 vế với 2 ta có

\(a^4+b^4\ge2ab\left(a^2+b^2\right)-2a^2b^2\)

\(\Leftrightarrow\left(a^2+b^2\right)^2\ge2ab\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)

Dẫu = xảy ra khi \(a=b\)

dat a+b=x b+c=y c+a=z \(\Rightarrow\) dt tro thanh \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\) \(\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\) (bdt amgm)

tuong tu \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\) \(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

\(\frac{\Rightarrow1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}.2\sqrt{\frac{xz}{\left(z+1\right)\left(x+1\right)}}.2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

                =\(8.\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}\)dau = xay ra khi x=y=z=1/2 hay a=b=c=1/4

Bạn xem lại, làm gì có cái ảnh đề nào đâu?

\(GT\Leftrightarrow\frac{1}{1+a}-1+\frac{1}{1+b}-1+\frac{1}{1+c}-1+\frac{1}{1+d}-1\)\(\ge3-4\)

\(\Rightarrow\frac{-a}{1+a}+\frac{-b}{1+b}+\frac{-c}{1+c}+\frac{-d}{1+d}\ge-1\)

\(\Rightarrow\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\le1\)

\(\Rightarrow\frac{a\left(1+b\right)+b\left(1+a\right)}{\left(1+a\right)\left(1+b\right)}+\frac{c\left(1+d\right)+d\left(1+c\right)}{\left(1+c\right)\left(1+d\right)}\le1\)

\(\Rightarrow\frac{a+2ab+b}{1+a+b+ab}+\frac{c+2cd+d}{1+c+d+cd}\le1\)

Áp dụng BĐT Cô - si , ta có:

\(1\ge\frac{2\sqrt{ab}+2ab}{1+2\sqrt{ab}+ab}+\frac{2\sqrt{cd}+2cd}{1+2\sqrt{cd}+cd}=\frac{2\sqrt{ab}}{1+\sqrt{ab}}+\frac{2\sqrt{cd}}{1+\sqrt{cd}}\)

\(\Rightarrow1\ge2\left[2\sqrt{\frac{\sqrt{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}}\right]\)\(=4.\frac{\sqrt[4]{abcd}}{1+\sqrt{ab}+\sqrt{cd}+\sqrt{abcd}}\)

\(\Rightarrow1\ge\frac{4\sqrt[4]{abcd}}{1+2\sqrt[4]{abcd}+\sqrt{abcd}}=\frac{4\sqrt[4]{abcd}}{\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}}\)

\(\Rightarrow4\sqrt[4]{abcd}\le\sqrt{\left(1+\sqrt[4]{abcd}\right)^2}\)

\(\Rightarrow4\sqrt[4]{abcd}\le1+\sqrt[4]{abcd}\)(vì a,b,c,d dương)

\(\Rightarrow3\sqrt[4]{abcd}\le1\)

\(\Rightarrow\sqrt[4]{abcd}\le\frac{1}{3}\)

\(\Rightarrow abcd\le\frac{1}{81}\)

(Dấu "="\(\Leftrightarrow a=b=c=d=\frac{1}{3}\))

Coll boy ! Bài này dòng 5 em áp dụng bất đẳng thức cô-si như vậy là chưa đúng nhé! Em kiểm tra lại mẫu trái dấu em nhé!

Nguyễn Linh Chia,b,c,d dương mà cô. Cô có thể nói cho em hiểu không ạ, chưa thông lắm??? )):

áp dụng bất đẳng thức Cauchy-schwaz

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}\)=\(\frac{16}{a+b+c+d}\)(đpcm)