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Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương:
\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)
Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
Áp dụng BĐT Cauchy-Schwarz ta có
BT\(\ge\)\(\frac{\left(1+1+1\right)^2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}=\frac{9}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\)
\(=\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}+\frac{7}{ab+bc+ac}\)
\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}+\frac{7}{ab+bc+ac}\)\(=1+\frac{7}{ab+bc+ac}\)
Ta lại có ab+bc+ac =< (a+b+c)^2/3 =3
\(\Rightarrow BT\ge1+\frac{7}{3}=\frac{10}{3}\)
Vậy GTNN là \(\frac{10}{3}\)khi a=b=c=1
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{1}{1+ab}+\frac{a^2}{a+ab}+\frac{b^2}{b+ab}\geq \frac{(1+a+b)^2}{1+ab+a+ab+b+ab}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a+b+1)^2}{a+b+1+3ab}\)
\(\Leftrightarrow \text{VT}\geq \frac{(a+b+1)^2}{a+b+1+3(3-a-b)}=\frac{(a+b+1)^2}{10-2(a+b)}\)
Theo giả thiết:
\(3=a+b+ab\Leftrightarrow 4=a+b+ab+1=(a+1)(b+1)\)
\(\leq \left (\frac{a+b+2}{2}\right)^2\) (theo BĐT AM-GM)
suy ra \(a+b+2\geq 4\Leftrightarrow a+b\geq 2\) (với \(a,b>0\) )
Do đó: \((a+b+1)^2\geq 9\) (1)
\(10-2(a+b)\leq 10-2.3=4; 10-2(a+b)=4+2ab>0\)
\(\Rightarrow \frac{1}{10-2(a+b)}\geq \frac{1}{6}\) (2)
Từ \((1);(2)\Rightarrow A\geq \frac{(a+b+1)^2}{10-2(a+b)}\geq \frac{9}{6}=\frac{3}{2}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=1\)
Hong Ra On Cái đó là BĐT Cauchy này nè :
\(xy\le\left(\dfrac{x+y}{2}\right)^2=\dfrac{\left(x+y\right)^2}{4}\)
Áp dụng vào:
\(\left(a+1\right)\left(b+1\right)\le\dfrac{\left(a+b+1+1\right)^2}{4}=\left(\dfrac{a+b+2}{2}\right)^2\)
a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)
\(=a-1\)
b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)
c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)