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Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
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
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Lời giải:
Đặt \(\left\{\begin{matrix} a+b-c=x\\ b+c-a=y\\ c+a-b=z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} b=\frac{x+y}{2}\\ c=\frac{y+z}{2}\\ a=\frac{x+z}{2}\end{matrix}\right.\) \((x,y,z>0\) do $a,b,c$ là ba cạnh tam giác ).
BĐT cần chứng minh tương đương với :
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)
Áp dụng BĐT Cauchy:
\(\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{2}{xy}\)
\(\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Theo BĐT S.Vacso: \(\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \frac{16}{(x+y)^2}(*)\)
Hoàn toàn tương tự:
\(2\left(\frac{1}{y^2}+\frac{1}{z^2}\right)\geq \frac{16}{(y+z)^2}; 2\left(\frac{1}{z^2}+\frac{1}{x^2}\right)\geq \frac{16}{(z+x)^2}(**)\)
Cộng theo vế \((*); (**)\) và rút gọn suy ra:
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z$ hay $a=b=c$
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
ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có
\(\Sigma\dfrac{1}{\left(a+b\right)c}\le\Sigma\dfrac{1}{2\sqrt{c}}=\dfrac{\Sigma\sqrt{ab}}{2}\le\dfrac{\Sigma a}{2}\)(đpcm)