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Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
2)đk: x>=0 \(\frac{x+8}{\sqrt{x}+1}=\frac{x-1+9}{\sqrt{x}+1}=\frac{\left(\sqrt{x}-1\left(\sqrt{x}+1\right)\right)}{\sqrt{x}+1}+\frac{9}{\sqrt{x}+1}=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-2\)
\(x\ge0\Leftrightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\frac{9}{\sqrt{x}+1}>0\). áp dụng bđt cosi cho 2 số dương \(\sqrt{x}+1;\frac{9}{\sqrt{x}+1}\) ta có:
\(\sqrt{x}+1+\frac{9}{\sqrt{x}+1}\ge2\sqrt{9}=6\Leftrightarrow\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-2\ge6-2=4\)=> Min =4 <=> x=4.
nhớ l i k e
:3 em từ olm sang đây có gì sai thì chỉ bảo
Áp dụng bất đẳng thức \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\forall x;y;z\inℝ\)
ta có \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc>0\Rightarrow ab+bc+ca\ge3\sqrt{abc}\)Ta lại có \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\forall a;b;c>0\)
Thật vậy \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)
\(\ge1+3\sqrt[3]{abc}+3\sqrt[3]{\left(abc\right)^2}+abc=\left(1+\sqrt[3]{abc}\right)^3\)
Khi đó \(P\le\frac{2}{3\left(1+\sqrt{abc}\right)}+\frac{\sqrt[3]{abc}}{1+\sqrt[3]{abc}}+\frac{\sqrt{abc}}{6}\)
Đặt \(\sqrt[6]{abc}=t\Rightarrow\sqrt[3]{abc}=t^2,\sqrt{abc}=t^3\)
Vì a,b,c > 0 nên 0<abc \(\le\left(\frac{a+b+c}{3}\right)^2=1\Rightarrow0< t\le1\)
Xét hàm số \(f\left(t\right)=\frac{2}{3\left(1+t^3\right)}+\frac{t^2}{1+t^2}+\frac{1}{6}t^3;t\in(0;1]\)
\(\Rightarrow f'\left(t\right)=\frac{2t\left(t-1\right)\left(t^5-1\right)}{\left(1+t^3\right)^2\left(1+t^2\right)^2}+\frac{1}{2}t^2>0\forall t\in(0;1]\)
Do hàm số đồng biến trên (0;1] nên \(f\left(t\right)< f\left(1\right)\Rightarrow P\le1\)
\(\Rightarrow\frac{2}{3+ab+bc+ca}+\frac{\sqrt{abc}}{6}+\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\le1\)
Dấu ''='' xảy ra khi \(a=b=c=1\)
Bài 2 :
Ta có :
\(2a^2+16ab+7b^2=\left(2a+3b\right)^2-2\left(a-b\right)^2\le\left(2a+3b\right)^2\)
\(\Rightarrow P\ge\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(a+3\right)}{a}\)
Áp dụng BĐT Cô - si ta có :
\(\frac{25a^2}{2a+3b}+2a+3b\ge10a\)
\(\frac{25b^2}{2b+3c}+2b+3c\ge10b\)
\(\frac{c^2\left(a+3\right)}{a}=\left(c^2+1\right)+\left(\frac{3c^2}{a}+3a\right)-3a-1\ge2c+6c-3a-1=8c-3a-1\)
Khi đó :
\(P\ge\left(10-2a-3b\right)+\left(10b-2b-3c\right)+\left(8c-3a-1\right)\)
\(\Rightarrow P\ge5\left(a+b+c\right)-1=14\)
Vậy \(MinP=14\) khi a=b=c=1
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
3, \(P=a+b+\frac{1}{2a}+\frac{2}{b}\)
=\(\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\)
AD bđt cosi vs hai số dương có:
\(\frac{1}{2a}+\frac{a}{2}\ge2\sqrt{\frac{1}{2a}.\frac{a}{2}}=2\sqrt{\frac{1}{4}}=1\)
\(\frac{b}{2}+\frac{2}{b}\ge2\sqrt{\frac{b}{2}.\frac{2}{b}}=2\)
Có \(\frac{a+b}{2}\ge\frac{3}{2}\) (vì a+b \(\ge3\))
=> \(P=\left(\frac{1}{2a}+\frac{a}{2}\right)+\left(\frac{b}{2}+\frac{2}{b}\right)+\frac{a+b}{2}\ge1+2+\frac{3}{2}\)
<=> P \(\ge4.5\)
Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}\frac{1}{2a}=\frac{a}{2}\\\frac{b}{2}=\frac{2}{b}\\a+b=3\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a^2=1\\b^2=4\\a+b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=1\\b=2\\a+b=3\end{matrix}\right.\)
=> a=2,b=3
Vậy minP=4.5 <=>a=1,b=2
a/
-Cauchy-Schwar
\(P=\sum\frac{a^4}{a\sqrt{b^2+3}}\ge\frac{\left(\sum a^2\right)^2}{\sum a\sqrt{b^2+3}}\)
Côsi: \(\sum a\sqrt{b^2+3}=\frac{1}{2}\sum2a.\sqrt{b^2+3}\le\frac{1}{2}.\sum\frac{\left(2a\right)^2+b^2+3}{2}=\frac{1}{4}.\left[5\left(a^2+b^2+c^2\right)+3.3\right]=6\)
\(\Rightarrow P\ge\frac{3^2}{6}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1.
b/
Côsi: \(8^x+8^x+64\ge3\sqrt[3]{8^x.8^x.64}=12.4^x\Rightarrow8^x\ge6.4^x-32\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-96\)
\(4^x+4^y+4^z\ge3\sqrt[3]{4^{x+y+z}}=3\sqrt[3]{4^6}=48\)
\(\Rightarrow-2\left(4^x+4^y+4^z\right)\le-96\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-2\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)