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Ta có: \(a+b+c+\sqrt{abc}=4\)
\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)
\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)
\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)
Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)
Lời giải:
Áp dụng BĐT AM-GM:
\(a^3+1=(a+1)(a^2-a+1)\leq \left(\frac{a+1+a^2-a+1}{2}\right)^2=\left(\frac{a^2+2}{2}\right)^2\)
\(b^3+1\leq \left(\frac{b^2+2}{2}\right)^2\)
\(\Rightarrow \sqrt{(a^3+1)(b^3+1)}\leq \frac{(a^2+2)(b^2+2)}{4}\)
\(\Rightarrow \frac{a^2}{\sqrt{(a^3+1)(b^3+1)}}\geq \frac{4a^2}{(a^2+2)(b^2+2)}\)
Hoàn toàn tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\geq \underbrace{\frac{4a^2}{(a^2+2)(b^2+2)}+\frac{4b^2}{(b^2+2)(c^2+2)}+\frac{4c^2}{(c^2+2)(a^2+2)}}_{M}\)
Ta cần CM \(M\geq \frac{4}{3}\)
\(\Leftrightarrow \frac{a^2(c^2+2)+b^2(a^2+2)+c^2(b^2+2)}{(a^2+2)(b^2+2)(c^2+2)}\geq \frac{1}{3}\)
\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (a^2+2)(b^2+2)(c^2+2)\)
\(\Leftrightarrow 3(a^2b^2+b^2c^2+c^2a^2)+6(a^2+b^2+c^2)\geq (abc)^2+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2+b^2+c^2)+8\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2(a^2+b^2+c^2)\geq 72\)
Điều này luôn đúng do theo BĐT AM-GM thì: \(\left\{\begin{matrix} a^2b^2+b^2c^2+c^2a^2\geq 3\sqrt[3]{(abc)^4}=3\sqrt[3]{8^4}=48\\ 2(a^2+b^2+c^2)\geq 6\sqrt[3]{(abc)^2}=6\sqrt[3]{8^2}=24\end{matrix}\right.\)
Do đó ta có đpcm
Dấu "=" xảy ra khi $a=b=c=2$
Bài 1 :
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=\left|x-1\right|=1-x\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=\left|y-1\right|=1-y\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=\left|z-1\right|=1-z\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
1) \(\frac{9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}=1\left(ĐK:x\ne0\right)\)
Đặt: \(\sqrt{2x^2+9}=a\left(a\ge0\right)\)
\(\Leftrightarrow2x^2+9=a^2\Leftrightarrow9=a^2-2a^2\)
Khi đó pt đã cgo trở rhanhf:
\(\frac{a^2-2x^2}{x^2}+\frac{2x}{a}=1\)
\(\Leftrightarrow\left(\frac{a}{x}\right)^2-2+\frac{2x}{a}-1=0\)
\(\Leftrightarrow\left(\frac{a}{x}\right)^2+\frac{2x}{a}-3=0\) (*)
Đặt: \(\frac{a}{x}=b\) khi đó (*) trở thành:
\(b^2+\frac{2}{b}-3=0\)
\(\Leftrightarrow b^3+2-3b=0\)
\(\Leftrightarrow\left(b^3-b\right)-\left(2b-2\right)=0\)
\(\Leftrightarrow b\left(b-1\right)\left(b+1\right)-2\left(b-1\right)=0\)
\(\Leftrightarrow\left(b-1\right)\left(b^2+b-2\right)=0\)
\(\Leftrightarrow\left(b-1\right)^2\left(b+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}b-1=0\\b+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}b=1\\b=-2\end{array}\right.\)
Với: \(b=1\) ta có:
\(\frac{a}{x}=1\Leftrightarrow a=x\Leftrightarrow\sqrt{2x^2+9}=x\Leftrightarrow2x^2+9=x^2\Leftrightarrow x^2+9=0\left(loai\right)\)
Với: \(b=-2\) ta có:
\(\frac{a}{x}=-2\)
\(\Leftrightarrow a=-2x\)
\(\Leftrightarrow\sqrt{2x^2+9}=-2x\)
\(\Leftrightarrow2x^2+9=4x^2\)
\(\Leftrightarrow2x^2=9\)
\(\Leftrightarrow x^2=\frac{9}{2}\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{\sqrt{2}}\\x=-\frac{3}{\sqrt{2}}\end{array}\right.\)
Thử lại ta thấy: \(x=\frac{3}{\sqrt{2}}\left(ktm\right);x=-\frac{3}{\sqrt{x}}\left(tm\right)\)
Vaayk pt đã cho có nhgieemj là \(x=-\frac{3}{\sqrt{2}}\)
cảm ơn bạn nhìu