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\(1=ab+bc+ca\le a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=\left(a+b+c\right)^2-2\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^2\ge3\)\(\Leftrightarrow\)\(a+b+c\ge\sqrt{3}\)
\(P=\frac{1}{1-\left(a+b\right)^2}+\frac{1}{1-\left(b+c\right)^2}+\frac{1}{1-\left(c+a\right)^2}\ge\frac{9}{3-\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]}\)
\(\ge\frac{9}{3-\frac{\left(2a+2b+2c\right)^2}{3}}=\frac{9}{3-\frac{4\left(a+b+c\right)^2}{3}}\ge\frac{9}{3-\frac{4.\left(\sqrt{3}\right)^2}{3}}=\frac{9}{3-4}=\frac{9}{-1}=-9\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{\sqrt{3}}\)
\(a^2+b^2+c^2\ge ab+ac+bc=1\)
\(P=\frac{1}{1-\left(a+b\right)^2}+\frac{1}{1-\left(b+c\right)^2}+\frac{1}{1-\left(c+a\right)^2}\)
\(P\ge\frac{9}{3-\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]}=\frac{9}{3-2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)}\)
\(\ge\frac{9}{3-2-2}=-9\)
Dấu "=" xảy ra <=> a=b=c
ab + ac + bc =1 <=> 3a^2 = 1 <=> a^2 = 1/3 \(\Rightarrow a=\frac{\sqrt{3}}{3}\) (Do a dương)
\(\Rightarrow b=c=a=\frac{\sqrt{3}}{3}\)
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có:
\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)
\(\ge\left(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\) \(\left(1\right)\)
Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\) ( Do abc=1 )
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=1\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)
Mà \(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\) (đpcm)
1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
Từ giả thiết \(ab+bc+ca=2abc\)suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\)thì \(\hept{\begin{cases}x+y+z=2\\x,y,z>0\end{cases}}\)và bất đẳng thức cần chứng minh trở thành \(\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge\frac{1}{2}\)hay \(\frac{x^3}{\left(y+z\right)^2}+\frac{y^3}{\left(z+x\right)^2}+\frac{z^3}{\left(x+y\right)^2}\ge\frac{1}{2}\)
Áp dụng bất đẳng thức Bunyakovsky dạng phân thức ta được \(\frac{x^3}{\left(y+z\right)^2}+\frac{y^3}{\left(z+x\right)^2}+\frac{z^3}{\left(x+y\right)^2}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2}\)\(=\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz}\)
Ta cần chứng minh\(\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz}\ge\frac{1}{2}\)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)^2\ge x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+6xyz\)
Thật vậy, theo một đánh giá quen thuộc ta có \(2\left(x^2+y^2+z^2\right)^2=2\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)\)\(\ge\frac{2\left(x+y+z\right)^2\left(x^2+y^2+z^2\right)}{3}\)
Mà ta lại có \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)=x^3+y^3+z^3+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\)
Suy ra ta có \(\frac{2\left(x+y+z\right)^2\left(x^2+y^2+z^2\right)}{3}\ge\frac{4\left(x^3+y^3+z^3+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\right)}{3}\)
Ta cần chỉ ra được \(4\left(x^3+y^3+z^3+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\right)\)\(\ge3\left(x^2y+y^2x+x^2z+z^2x+y^2z+yz^2+6xyz\right)\)
Hay\(4\left(x^3+y^3+z^3\right)+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\ge18xyz\)
Áp dụng bất đẳng thức Cauchy ta được \(4\left(x^3+y^3+z^3\right)\ge12xyz\); \(x^2y+y^2z+z^2x\ge3xyz\); \(xy^2+yz^2+zx^2\ge3xyz\)
Cộng theo vế các bất đẳng thức trên ta được\(4\left(x^3+y^3+z^3\right)+x^2y+y^2x+x^2z+z^2x+y^2z+yz^2\ge18xyz\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{3}{2}\)
\(a+b+c=1\)
\(\Rightarrow\left(a+b+c\right)^2=1\)
\(\Rightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1-2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2=1-2\left(ab+bc+ca\right)\)
Lại có:
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow abc\le\frac{ab+bc+ca}{9}\)
Khi đó:
\(M\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=21+9=30\)
Dấu "=" xảy ra tại \(a=b=c=\frac{1}{3}\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
Đặt \(A=\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ca}=\frac{a^2+2ab+b^2}{ab}+\frac{b^2+2bc+c^2}{bc}+\frac{c^2+2ac+c^2}{ca}\)
\(=\frac{a}{b}+2+\frac{b}{a}+\frac{b}{c}+2+\frac{c}{b}+\frac{c}{a}+2+\frac{a}{c}=6+a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{b}+\frac{1}{a}\right)\)
\(\ge6+\frac{4a}{b+c}+\frac{4b}{c+a}+\frac{4c}{a+b}\ge6+2\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+b}\right)+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
\(\ge6+2\cdot\frac{3}{2}+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=9+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
Dấu "=" xảy ra <=> a=b=c