$89-$ Let $a, b, c, d, e$ be non-negative real numbers, and $a+b=c+d+e$, find the largest positive real number $\bar{T}$, such that the inequality $\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}} \geqslant T(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2}$ holds. (2007 Iran Mathematical Olympiad)
89. Taking $a=b=3, c=d=e=2$, we get $T \leqslant \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}$.

Next, we prove
$$\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}} \geqslant \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2}$$

Let $X=a+b=c+d+e$, by the Cauchy-Schwarz inequality we have $a^{2}+b^{2} \geqslant \frac{(a+b)^{2}}{2}=\frac{X^{2}}{2}, c^{2}+d^{2}+e^{2} \geqslant \frac{(c+d+e)^{2}}{3}=\frac{X^{2}}{3}$, adding these gives
$$a^{2}+b^{2}+c^{2}+d^{2}+e^{2} \geqslant \frac{5 X^{2}}{6}$$

Again by the Cauchy-Schwarz inequality, we have $\sqrt{a}+\sqrt{b} \leqslant \sqrt{2(a+b)}=\sqrt{2 X}, \sqrt{c}+\sqrt{d}+\sqrt{e} \leqslant \sqrt{3(c+d+e)}=\sqrt{3 X}$, adding these gives
$$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e} \leqslant(\sqrt{3}+\sqrt{2}) X$$

Thus,
$$(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2} \leqslant(\sqrt{3}+\sqrt{2})^{2} X^{2}$$

From (1) and (2) we get
$$\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}} \geqslant \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2}$$

Equality holds if and only if $\frac{2 a}{3}=\frac{2 b}{3}=c=d=e$.