Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)

Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\) 

Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\) 

" = " \(\Leftrightarrow a=b=c=1\)

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1

Không mất tính tổng quát, giả sử đó là a và b

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)

\(\Leftrightarrow ab+1\ge a+b\)

\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)

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

Lại có:

\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)

\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)

\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

chết đăng nhầm sogy nha

1.

Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)

\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)

\(\RightarrowĐpcm\)

2.

Đành giải tạm bằng nick này vì sợ một vài thành phần trẻ trâu anti phá phách :poor:

Phân tích và giải

Dễ thấy: Dấu "=" khi \(a=b=c=1\)

\(\Rightarrow L=Σ\dfrac{a}{\left(a+1\right)^2}=\dfrac{3}{4}\text{ và }F=-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=-\dfrac{1}{2}\)

Khi đó \(VT=L-F=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)

Ta sẽ chia làm 2 bước cm:

B1: \(Σ\dfrac{a}{\left(a+1\right)^2}\le\dfrac{3}{4}\). Ta xét BĐT :

\(\dfrac{a}{\left(a+1\right)^2}=\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2k}+a^k\right)}{8\left(a^{2k}+a^k+1\right)}\) (cần tìm \(k\) thỏa mãn)

\(\Leftrightarrow8a\left(a^{2k}+a^k+1\right)-3\left(a^{2k}+a^k\right)\left(a^2+2a+1\right)\le0\)\(\Leftrightarrow f\left(a\right)=-3a^{2k}+2a^{k+1}-3a^{k+2}+2a^{2k+1}-3a^{2k+2}-3a^k+8a\)

\(\Rightarrow f'\left(a\right)=2k\cdot-3a^{2k-1}+\left(k+1\right)2a^k-\left(k+2\right)3a^{k+1}+\left(2k+1\right)2a^{2k}-\left(2k+2\right)3a^{2k+1}-k\cdot3a^{k-1}+8a\)

\(\Rightarrow f'\left(1\right)=0\Rightarrow-12k=0\Rightarrow k=0\)

Hay BĐT phụ cần tìm là \(\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2\cdot0}+a^0\right)}{8\left(a^{2\cdot0}+a^0+1\right)}=\dfrac{1}{4}\) (bài này \(k\) đẹp ra luôn \(\farac{1}{4}\) cộng vào là ok =))

\(\Leftrightarrow-\dfrac{\left(a-1\right)^2}{4\left(a+1\right)^2}\le0\) *Đúng* \(\RightarrowΣ\dfrac{a}{\left(a+1\right)^2}\leΣ\dfrac{1}{4}=\dfrac{3}{4}\)

B2: CM \(-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le-\dfrac{1}{2}\)

Tự cm nhé Goodluck :v

đây là hệ số bất định

Một lời giải sơ cấp:

Đổi \(\left(a;b;c\right)\rightarrow\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\).BDT cần chứng minh tương đương:

\(\sum\dfrac{xy}{\left(x+y\right)^2}-\dfrac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\dfrac{1}{4}\)

\(\Leftrightarrow\left[\dfrac{3}{4}-\sum\dfrac{xy}{\left(x+y\right)^2}\right]+\left[\dfrac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}-\dfrac{1}{2}\right]\ge0\)

\(\Leftrightarrow\sum\left[\dfrac{1}{4}-\dfrac{xy}{\left(x+y\right)^2}\right]-\dfrac{\sum\left(x^2+y^2\right)z-6xyz}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(x-y\right)^2}{4\left(x+y\right)^2}-\dfrac{\sum z\left(x-y\right)^2}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge0\)

\(\Leftrightarrow\sum\left(x-y\right)^2\left[\dfrac{1}{4\left(x+y\right)^2}-\dfrac{z}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\right]\ge0\)

hay \(S_a\left(y-z\right)^2+S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\ge0\)(*)

với \(\left\{{}\begin{matrix}S_a=\dfrac{1}{4\left(y+z\right)^2}-\dfrac{x}{2\prod\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-z\right)}{4\left(y+z\right)^2\left(x+y\right)\left(x+z\right)}\\S_b=\dfrac{1}{4\left(x+z\right)^2}-\dfrac{y}{2\prod\left(x+y\right)}=\dfrac{\left(y-x\right)\left(y-z\right)}{4\left(x+z\right)^2\left(x+y\right)\left(y+z\right)}\\S_c=\dfrac{1}{4\left(x+y\right)^2}-\dfrac{z}{2\prod\left(x+y\right)}=\dfrac{\left(z-x\right)\left(z-y\right)}{4\left(x+y\right)^2\left(y+z\right)\left(z+x\right)}\end{matrix}\right.\)

Dễ thấy \(S_a;S_b;S_c\) không phải là luôn không âm.Giả sử \(x=max\left\{x;y;z\right\}\).

Từ đó suy ra \(S_a\ge0\).Xét \(S_b+S_c=\dfrac{\left(y-z\right)^2}{4\left(x+y\right)^2\left(x+z\right)^2}\ge0,\forall x;y;z>0\)

Do đó \(VT=S_a\left(x-y\right)^2+\left[S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\right]\ge0\)

Ta sẽ chứng minh \(S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\ge0\) với \(S_b+S_c\ge0\)

và điều này đúng hay không e không biết, quan trọng là .. Chúc Mừng Năm Mới !!

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.