Gọi tam giác đó là ABC, đường cao AH. (Bạn tự vẽ hình nhá)

Xét ΔABH và ΔACH có AB = AC, ∠AHB = ∠AHC = 90, chung cạnh AH.Dó đó ΔABH = ΔACH (ch.cgv)

=> \(HB=HC=\dfrac{BC}{2}=\dfrac{a}{2}\)

Vì ΔABH vuông tại H, áp dụng định lí Py - ta - go, ta có \(AH^2+HB^2=AB^2\Rightarrow AH^2=a^2-\dfrac{a^2}{4}=\dfrac{3}{4}a^2\Rightarrow AH=\dfrac{\sqrt{3}}{2}a\)

Áp dụng BĐT AM - GM, ta có \(\sqrt{\dfrac{2}{3}}P=\sqrt{\dfrac{2}{3}\left(a+b\right)}+\sqrt{\dfrac{2}{3}\left(b+c\right)}+\sqrt{\dfrac{2}{3}\left(c+a\right)}\le\dfrac{\dfrac{2}{3}+a+b}{2}+\dfrac{\dfrac{2}{3}+b+c}{2}+\dfrac{\dfrac{2}{3}+c+a}{2}=\dfrac{2+2\left(a+b+c\right)}{2}=2\Rightarrow P\le\sqrt{6}\)Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{3}\)

Để em xem xét :)

Áp dụng BĐT AM - GM, ta có \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{64\left(1+y\right)\left(1+z\right)}}=\dfrac{3}{4}x\)

Chứng minh tương tự, ta có \(A+\dfrac{1+x}{4}+\dfrac{1+y}{4}+\dfrac{1+z}{4}\ge\dfrac{3}{4}\left(x+y+z\right)\Rightarrow A\ge\dfrac{1}{2}\left(x+y+z\right)-\dfrac{3}{4}\)

Áp dụng BĐT AM - GM, ta có \(x+y+z\ge3\sqrt[3]{xyz}=3\)

Do đó \(A\ge\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)

Dấu = xảy ra <=> x = y = z = 1.

Ta chứng minh BĐT phụ \(x^3+y^3\ge xy\left(x+y\right)\left(x>0,y>0\right)\Leftrightarrow x^2-xy+y^2\ge xy\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Áp dụng BĐT, ta được \(A=\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}+\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}=\dfrac{1}{ab\left(a+b+c\right)}+\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}=\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\left(đpcm\right)\)

Dấu = xảy ra <=> a = b = c.

Ta chứng minh \(\dfrac{1}{\left(k+1\right)\sqrt{k}}+\dfrac{1}{k\sqrt{k+1}}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)

Thật vậy, ta có \(\dfrac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\dfrac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{\left(k+1\right)^2k-k^2\left(k+1\right)}=\dfrac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{\left(k+1\right)k}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)(luôn đúng)

Khi đó, ta có \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2009}}-\dfrac{1}{\sqrt{2010}}=1-\dfrac{1}{\sqrt{2010}}=\dfrac{\sqrt{2010}-1}{\sqrt{2010}}\)