ca này để thầy lâm ròi:<

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

\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}=\dfrac{5-a}{16}\)

Tương tự: \(\dfrac{1}{b^2+c+d+a}\le\dfrac{5-b}{16}\) ...

Cộng vế:

\(P\le\dfrac{20-\left(a+b+c+d\right)}{16}=1\)

Dấu "=" xảy ra khi \(a=b=c=d=1\)

Bunhiacopxki:

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

\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}\)

Tương tự:

\(\dfrac{1}{b^2+c+d+a}\le\dfrac{1+c+d+a}{16}\) ; \(\dfrac{1}{c^2+d+a+b}\le\dfrac{1+d+a+b}{16}\)

\(\dfrac{1}{d^2+a+b+c}\le\dfrac{1+a+b+c}{16}\)

Cộng vế:

\(P\le\dfrac{4+3\left(a+b+c+d\right)}{16}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=d=1\)

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

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

Áp dụng BĐT Svacxơ:

\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\ge\dfrac{4}{ab+bc+cd+da}\)

Áp dụng BĐT Cô-si:

\(\dfrac{4}{ab+bc+cd+da}\ge\dfrac{4}{a^2+b^2+c^2+d^2}\)

Ta cần c/m: \(\dfrac{4}{a^2+b^2+c^2+d^2}\ge a^2+b^2+c^2+d^2\)

\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)^2\ge4\)

Áp dụng BĐT Svacxơ: \(\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{c^2}{1}+\dfrac{d^2}{1}\right)^2\ge\dfrac{\left(a+b+c+d\right)^{2^2}}{16}\)

mà a+b+c+d=4 nên: \(\dfrac{\left(a+b+c+d\right)^4}{16}\ge\dfrac{64}{16}=4=VP\)

Vậy ta có đpcm.

a+b+c+d=4 nha

\(b+c\le\sqrt{2\left(b^2+c^2\right)}\Rightarrow\dfrac{a^2}{b+c}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{1}{\sqrt{2}}.\dfrac{a^2}{\sqrt{b^2+c^2}}\)

Sau đó làm tiếp như bài đó là được

Áp dụng cauchy-schwarz:

\(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+e}+\dfrac{d}{e+a}+\dfrac{e}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ce}+\dfrac{d^2}{ed+ad}+\dfrac{e^2}{ae+be}\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+ad+ae+bc+bd+be+cd+ce+de}\)

Giờ chỉ cần chứng minh

\(ab+ac+ad+ae+bc+bd+be+cd+ce+de\le\dfrac{2}{5}\left(a+b+c+d+e\right)^2\)

\(\Leftrightarrow ab+ac+ad+ae+bc+bd+be+cd+ce+de\le2\left(a^2+b^2+c^2+d^2+e^2\right)\)

điều này hiển nhiên đúng theo AM-GM:

\(ab\le\dfrac{a^2+b^2}{2};ac\le\dfrac{a^2+c^2}{2};ad\le\dfrac{a^2+d^2}{2}...\)

Cứ vậy ta thu được đpcm .Dấu = xảy ra khi a=b=c=d=e

P/s: : ]