Cho a, b, c là số thực dươn. Chứng minh bất đẳng thức:
\(\dfrac{1}{a\left(a^2+8ab\right)}+\dfrac{1}{b\left(b^2+8ac\right)}+\dfrac{1}{c\left(c^2+8ab\right)}\le\dfrac{1}{3abc}\)
\(\dfrac{1}{a\left(a^2+8bc\right)}+\dfrac{1}{b\left(b^2+8ac\right)}+\dfrac{1}{c\left(c^2+8ab\right)}\le\dfrac{1}{3abc}\)
P/S: Bài khá hay (theo cảm nhận)
Cho a, b, c là ba số thực dương. Chứng minh bất đẳng thức:
\(\frac{1}{a\left(a^2+8bc\right)}+\frac{1}{b\left(b^2+8ca\right)}+\frac{1}{c\left(c^2+8ab\right)}\le\frac{1}{3abc}\)
:D
\(\frac{1}{a\left(a^2+8bc\right)}+\frac{1}{b\left(b^2+8ca\right)}+\frac{1}{c\left(c^2+ab\right)}\le\frac{1}{3abc}\)
\(\Leftrightarrow\frac{1}{\frac{a^2}{bc}+8}+\frac{1}{\frac{b^2}{ca}+8}+\frac{1}{\frac{c^2}{ab}+8}\le3\) (*)
Đặt \(\frac{a^2}{bc}=x;\frac{b^2}{ca}=y;\frac{c^2}{ab}=z\left(x,y,z>0\right)\)
(*)\(\Leftrightarrow\frac{1}{x+8}+\frac{1}{y+8}+\frac{1}{z+8}\le\frac{1}{3}\)
\(\Leftrightarrow16\left(x+y+z\right)+5\left(xy+yz+zx\right)\ge63\)(**)
(**) đúng bởi \(x+y+z\ge3\sqrt[3]{xyz}=3;xy+yz+zx\ge3\sqrt[3]{\left(xyz\right)^2}=3\)
Cho ba số thực dương a,b,c thỏa mãn . Chứng mình rằng:
\(\left(a+b+c\right)+2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+\dfrac{8}{abc}\ge\dfrac{121}{12}\)
Tách biểu thức như sau:
\(\left(\dfrac{a}{9}+\dfrac{b}{12}+\dfrac{c}{6}+\dfrac{8}{abc}\right)+\left(\dfrac{a}{18}+\dfrac{b}{24}+\dfrac{2}{ab}\right)+\left(\dfrac{b}{16}+\dfrac{c}{8}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{c}{6}+\dfrac{2}{ca}\right)+\left(\dfrac{13a}{18}+\dfrac{13b}{24}\right)+\left(\dfrac{13b}{48}+\dfrac{13c}{24}\right)\)
Cho a, b, c, d > 0. Chứng minh rằng:
1.
\(\dfrac{a}{\sqrt{a^2+8bc}}\)+ \(\dfrac{b}{\sqrt{b^2+8ac}}\)+ \(\dfrac{c}{\sqrt{c^2+8ab}}\) ≥ 1
2.
\(\dfrac{a}{b+2c+3d}\)+\(\dfrac{b}{c+2d+3a}\)+\(\dfrac{c}{d+2a+3b}\)+ \(\dfrac{d}{a+2b+3c}\) ≥ \(\dfrac{2}{3}\)
3.
\(\dfrac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}\) + \(\dfrac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}\) + \(\dfrac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}\) + \(\dfrac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\) ≥ \(\dfrac{a+b+c+d}{4}\)
Bất đẳng thức BuNyaKovSky ( BCS )
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
Chứng minh bất đẳng thức sau:
\(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(a,b,c>0\right)\)
Áp dụng BĐT cosi:
\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)
Dấu \("="\Leftrightarrow a=b=c\)
Cho 3 số thực dương a,b,c thoả mãn : ab+bc+ca=3 .Chứng minh :
\(\dfrac{1}{1+a^2\left(b+c\right)}+\dfrac{1}{1+b^2\left(c+a\right)}+\dfrac{1}{1+c^2\left(a+b\right)}\le\dfrac{1}{abc}\)
\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)
\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)
a, Cho a,b là số thực dương và ab<1. Chứng minh \(\dfrac{1}{1+a}+\dfrac{1}{1+b}\le\dfrac{2}{1+\sqrt{ab}}\)
b, Cho a,b,c là các số thực dương thõa mãn abc=1. Chứng minh \(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)
Cho a,b,c dương. Chứng minh
\(\dfrac{1}{\left(a+b\right)^2}+\dfrac{1}{\left(b+c\right)^2}+\dfrac{1}{\left(c+a\right)^2}\ge\dfrac{3\sqrt{3abc\left(a+b+c\right)}.\left(a+b+c\right)^2}{4\left(ab+bc+ca\right)^3}\)
Cho a,b,c>0 thỏa abc=1. Chứng minh :
\(\dfrac{a}{\left(a+1\right)^2}+\dfrac{b}{\left(b+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le\dfrac{1}{4}\)
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
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 !!