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Ta chứng minh BĐT sau cho các số dương:
\(x^5+y^5\ge xy\left(x^3+y^3\right)\)
\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)
\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)
Áp dụng:
\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)
Tương tự và cộng lại:
\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)
\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)
\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)
Tham khảo:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
Lời giải khác:
Theo BĐT AM-GM:
\(\text{VT}=\sum \frac{\sqrt{2(b^2+c^2)-a^2}}{a}\geq \sum \frac{\sqrt{(b+c)^2-a^2}}{a}=\sum \frac{\sqrt{a+b+c}.\sqrt{b+c-a}}{a}\)
\(=\sum \frac{\sqrt{a+b+c}.(b+c-a)}{\sqrt{a^2(b+c-a)}}\)
Theo BĐT AM-GM:
$a^2(b+c-a)\leq \left(\frac{a+b+c}{3}\right)^3$
\(\Rightarrow \text{VT}\geq 3\sqrt{3}\sum \frac{\sqrt{a+b+c}(b+c-a)}{\sqrt{(a+b+c)^3}}=3\sqrt{3}.\sum \frac{b+c-a}{a+b+c}=3\sqrt{3}\)
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Chuẩn hóa \(a+b+c=3\)
Do a;b;c là độ dài 3 cạnh của 1 tam giác nên ta cũng suy ra \(0< a;b;c< \frac{3}{2}\)
Đặt vế trái là P, ta có:
\(P=\sum\frac{\sqrt{2\left(b^2+c^2\right)-a^2}}{a}\ge\sum\frac{\sqrt{\left(b+c\right)^2-a^2}}{a}=\sum\frac{\sqrt{\left(a+b+c\right)\left(b+c-a\right)}}{a}=\sqrt{3}\left(\frac{\sqrt{3-2a}}{a}+\frac{\sqrt{3-2b}}{b}+\frac{\sqrt{3-2c}}{c}\right)\)
Ta có đánh giá: \(\frac{\sqrt{3-2a}}{a}\ge3-2a\) với mọi \(a\in\left(0;\frac{3}{2}\right)\)
Thật vậy, BĐT \(\Leftrightarrow a\sqrt{3-2a}\le1\)
\(\Leftrightarrow1-a^2\left(3-2a\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(2a+1\right)\ge0\) (luôn đúng)
Tương tự \(\frac{\sqrt{3-2b}}{b}\ge3-2b\) ; \(\frac{\sqrt{3-2c}}{c}\ge3-2c\)
\(\Rightarrow P\ge\sqrt{3}\left[9-2\left(a+b+c\right)\right]=3\sqrt{3}\) (đpcm)
\(VT=\sqrt{\left(2+2a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(VT=\sqrt{\left[a^2-2a+1+a^2+2a+1\right]\left[b^2+2bc+c^2+b^2c^2-2bc+1\right]}\)
\(VT=\sqrt{\left[\left(1-a\right)^2+\left(a+1\right)^2\right]\left[\left(bc-1\right)^2+\left(b+c\right)^2\right]}\)
Bunhiacopxki:
\(VT\ge\left(1-a\right)\left(bc-1\right)+\left(a+1\right)\left(b+c\right)=\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\left(1+abc\right)\)
\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)
\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)
Tương tự và cộng lại:
\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
