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\(A=\frac{a\sqrt{a}}{\sqrt{a+b+2c}}+\frac{b\sqrt{b}}{\sqrt{b+c+2a}}+\frac{c\sqrt{c}}{\sqrt{c+a+2b}}\)
\(A=\frac{a^2}{\sqrt{a\left(a+b+2c\right)}}+\frac{b^2}{\sqrt{b\left(b+c+2a\right)}}+\frac{c^2}{\sqrt{c\left(c+a+2b\right)}}\)
\(\ge\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\)
Xét: \(2\left(\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\right)\)
\(=\sqrt{4a\left(a+b+2c\right)}+\sqrt{4b\left(b+c+2a\right)}+\sqrt{4c\left(c+a+2b\right)}\)
\(\le\frac{4a+a+b+2c+4b+b+c+2a+4c+c+a+2b}{2}=4\left(a+b+c\right)\)
\(\Rightarrow\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\le2\left(a+b+c\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
ÁP DỤNG BĐT COSI TA CÓ :\(\sqrt{\frac{a}{b+c+2a}}\le\frac{a}{b+c+2a}+\frac{1}{4}\)
\(\sqrt[]{\frac{b}{a+c+2b}}\le\frac{b}{a+c+2b}+\frac{1}{4}\)
\(\sqrt[]{\frac{c}{a+b+2c}}\le\frac{c}{a+b+2c}+\frac{1}{4}\)
ĐẶT A=\(\sqrt[]{\frac{a}{b+c+2a}}+\sqrt[]{\frac{b}{a+c+2b}}+\sqrt[]{\frac{c}{a+b+2c}}\)
\(\le\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}+\frac{3}{4}\)
ÁP DỤNG BĐT :\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Rightarrow\frac{a}{b+c+2a}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
\(\Rightarrow\frac{b}{a+c+2b}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(\Rightarrow\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{c+b}\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{1}{4}\left(1+1+1\right)+\frac{3}{4}\)
\(\Rightarrow A\le\frac{3}{2}\)
DẤU = XẢY RA\(\Leftrightarrow a=b=c\)
Một lời giải khác:
\(\left(\Sigma\sqrt{\frac{a}{b+c+2a}}\right)^2=\left(\Sigma\sqrt{\frac{a\left(a+2c+b\right)}{\left(a+2c+b\right)\left(b+c+2a\right)}}\right)^2\)
\(\le\left[\Sigma a\left(a+2c+b\right)\right]\left[\Sigma\frac{1}{\left(a+2c+b\right)\left(b+c+2a\right)}\right]=\Sigma\frac{a^2+3ab}{\left(a+2c+b\right)\left(b+c+2a\right)}\)
\(=\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\)
Cần chứng minh \(\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\le\frac{9}{4}\)
Chịu khó quy đồng :V
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
Lời giải:
Với $a,b,c>0$ dễ thấy $0< \frac{a}{a+2b}< 1$
$\Rightarrow 0< \sqrt{\frac{a}{a+2b}}< 1$
$\Rightarrow \sqrt{\frac{a}{a+2b}}> \frac{a}{a+2b}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
$\text{VT}> \frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}$
Áp dụng BĐT Cauchy-Schwarz:
$\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq \frac{(a+b+c)^2}{a^2+2ba+b^2+2cb+c^2+2ac}=1$
Do đó $\text{VT}>1$ (đpcm)
Sử dụng BĐT AM-GM:
\(VT=\sum\limits_{cyc} \sqrt{\frac{a}{a+2b}} =\sum\limits_{cyc} \frac{a}{\sqrt{a(a+2b}}\geq \sum\limits_{cyc} \frac{2a}{2(a+b)}\)
\(=\sum\limits_{cyc} \frac{a^2}{a^2 +ab} \ge \frac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} >\frac{(a+b+c)^2}{a^2+b^2+c^2+2ab+2bc+2ca} = 1\) (đpcm)
P/s: Em không chắc lắm.
Ez to prove \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)
\(\Leftrightarrow\frac{6054}{3}\ge ab+bc+ca\Leftrightarrow ab+ca+bc\le2018\)
Khi đó: \(\frac{2a}{\sqrt{a^2+2018}}\le\frac{2a}{\sqrt{a^2+ab+bc+ca}}=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+b}+\frac{a}{a+c}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(P\le\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}=3\)
Đặt A là biểu thức cần CM
ví dụ Từ ĐK a + b + c = 3 => a² + b² + c² ≥ 3 ( Tự chứng minh )
Áp dụng BĐT quen thuộc x² + y² ≥ 2xy
a^4 + b² ≥ 2a²b (1)
b^4 + c² ≥ 2b²c (2)
c^4 + a² ≥ 2c²a (3)
Trước hết ta chứng minh BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Thật vậy:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}.3\sqrt[3]{a.b.c}.3\sqrt[3]{ab.bc.ca}\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
Ta có:
\(A^2=\left(\sqrt{a+c}.\sqrt{\frac{2a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{a+b}.\sqrt{\frac{2b}{\left(a+b\right)\left(b+c\right)}}+\sqrt{b+c}\sqrt{\frac{2c}{\left(c+a\right)\left(b+c\right)}}\right)^2\)
\(\Rightarrow A^2\le\left(a+c+a+b+b+c\right)\left(\frac{2a}{\left(a+b\right)\left(a+c\right)}+\frac{2b}{\left(a+b\right)\left(b+c\right)}+\frac{2c}{\left(c+a\right)\left(b+c\right)}\right)\)
\(\Rightarrow A^2\le\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)}=9\)
\(\Rightarrow A\le3\)
\(A_{max}=3\) khi \(a=b=c\)